I did it this way: first found out a general term for S by observation as: 7*10^r+1 + 5*10^r-1+....+ 5*10+7 for a general 'r'. then just apply summation of this from r=0 to 98 and use gp formula on 10^r+1 and compare with the other form to get m and n. this is the most basic approach that can be thought of during exam i think, that 10S one is too intuitive for me.
@@Realsxullzsame bhai Maine bhi aise hi Kiya tough utna nahi tha isse 6 min me hogaya 10s wala method ko apply karne ke liye 13 ka table acha se yad hona cahiye tab hi wo click hoga ki Haan aise bhi kar sakte hain
Bhaiyon, 85 batch IITK. Kiya tha yeh sab ek zamane mein. Aisa laga ki kabhi kaam na aya per aise sawalon se dimag tez ho jaata hai - kahin bhi kisi se bhi compete kar paoge! Good luck to you all! IIT mein milenge!
What I understood from the question is that when the terms of the series are increasing digit by digit you should multiply by 10^(no of digit increases per term)
Well to each their own. Mujhe utna rattna pasand nhi 😂 pehle hum soche the neet de denge but accha hua maine biology time rehte chor diya 💀 11-12 ka bio mere bas ke bahaar hya; isse accha maths kar lunga, accha bhi lagta hya maths
*"Bruh why all the people in the comments are telling this was easy and even some are telling that they solved it orally (sarcasm)💀☠️. Bro ur whole comment is filled with toppers having AIR
aree ye sab fekte bohot hai. asli paper ka reality yahi h 50% marks bhi score nahi kar pate 99.8% of jee advanced aspirant that's a fact you cannot deny
Bhai ye Pathfinder ka sawal hai mai karta tha fun mein bachpan mein uska solved exam hai usme method tha 10 se multiply krdo aur subtracr krdo fir gp ban jayega 2 min mein ho gya mujhe fix approach pata tha aisa mere saath kash mere paper mein bhi ho jaye that's why reading solved example are a must
@@piyushjain7687 average jain 🛐🛐🛐 bhai jain aur agarwal kis mitti ke hote hai I can trust anyone easily mai toh tomar hu naam bhi nahi ata kisi ka top 1k mein bas achi baat ye hai I'm obc
@@Ayush-mg6xw majak kar Raha hu Mari preparation mei bss physics ke L Lage hai 😣 pathfinder try karne ke liye base hi nhi bann pa Raha hu maths badiya hai 🥵🗿bss iss questions ko try Kiya tha first attempt meaning bana tha because GP padhne ke direct baad iss question pe kud gaya tha
I was able to solve it in under 10 minutes using a much simpler and obvious approach. You can try adding up the unit, tens up till ten thousandth digit by adding 7*99 + (5*98 + 7)*10 + (5*97 + 7)*100 and you get 61683 as the last five digits. You know the sum must be of at least 100 digits so remainder “n” can not be greater than 12 otherwise the numerator would be greater than 101 digits which shouldn’t be possible since m and n are less than 3000. Then you can multiply the last 4 or 5 digits with each number up to 12 and will easily be able to note when you get to 9 that it’s the only time the multiple is greater than the sum in the threshold of 3000. You can continue to check a few more digits when multiplying by 9 but will see it easily and see the multiple ending in 6767 which is 1210 more than 5557 and you get 1219.
@@ManasChoudhary-p1xbruh at that point you’re just guessing the remainder to be 9 and hoping to be correct, it’s like saying why don’t you hit in the first trial of hit and trial
Another way of solving this question is by using method of difference which a lot of people don't really know about. If you notice, you'll find out that the common difference between consecutive terms is in GP. Therefore by using method of difference the general term can be given by a(r)^n-1 + b where 'a' and 'b' are variables, r is the common ratio of gp. Using n=1 ans n=2 you can find the value of a and b and thus you'll get the general term. Now just use sigma on general term and do some basic manipulation to get the answer. This is how I solved it during my first attempt. Edit - didn't see bhaiya already mentioned it in the video. I just commented without seeing the video
@@rawatutkarshBhai fir bhi sirf 424 bacche solve kiye . I mean ye out of the box ques bhi nhi tha only bhaiya ka method was like too out of the box . Actually me exam situation kuch level of difficulty badha deta hai or else Ghar me baithke toh kitne saari ne solve kar diya
I did break 77 = 22+55 757 = 202 + 555 7557 = 2002 + 5555 Now 2002 = 2(10^3 + 1) 5555 = (5/9)(10^4-1) And that's how I solved it, but it was still calculation expensive.
Sameer bansal definite integral Ex 5 ke Q.50 me ek result tha ki (111...11) n times= 10^(n) -1/9 waha se sochke pattern bana liya (755...57)=(777..77) (n+2) times - 10 x (222..22) n times, iske baad bas calculation aur consolidate karna tha.
sir mene i solved this today morning what i did was write 77 as 70+7 , 757 as 700+50+7 , 7557 as 7000 +500+50+7 .... which we will become s = 7*99 + 7*(10+100+1000....) + 50( 1+ 11 +111...) phir isko solve kara aur ansar agaya roughly 8min lage the
Same mene bhi Yahi socha tha pr mene aage ki calculation nhi ki kyuki mujhe maloom tha ki ye general approach h to answer aa hi jayega BTW well done !!❤
I solved it with a completely different approch jese bachpan me likhe the wese 77 + 757 +7557........ likha then isme pattern notice kiya ki is likh sakte hai as 7(99) [ones place ke sab 7 ka sum] +7(10+100+1000.....10^99) [Remaining 7s according to its place] +5(10+100+1000.....10^98) [Remaining 7s ke just right wale 5s] +5(10+100+1000.....10^97) [just right wale aur 5s] . . . 5(10) [aakheri tens place wala 5 jo series ka last digit hai] then (10+100+......) wale part me GP ke sum ka formula lagaya 75...99.....57 wala term dikh gya +57 -57 karna para kyuki initially 75............500 form hua tha then answer 1219 aa gya {First attempt me starting me 7(100)+......... [100 nhi 99 hona tha] kar diya tha ans 1282 aaya tha fir video dekha toh pata chala ans 1219 hai thora soochne ke baad galti dikh gai ki 7(99) +7(10+100+.......) hoga then correct answer aa gya }
Chutiya tarika hai yah 😂 bas calculation karta jao jao orh time waste hai orh kuch nhi . Rather than Asa karo S=77+757+7557+. N Multiple by 10 on both the side 10S=770+7570+75570+755570. 755..............570 Now subtract both then you will observe a pattern 9S=-77+13+13+13+13+13+13. N +7555555555.........70 Now 9S=13*98-77+7555555.......570 S=13*98-77+755555555.....70/9 Hence M+n is 1219@@4fgaming925
I was the only one in my coaching to have solved question like this alongside my teacher (obviously), in questions like these where some kind of recurrence pattern is forming, this AGP method works usually, a few terms will first trouble you but with enough practice it becomes easy So trust yourself and work hard
Can be done by writing expansion don't take that 99th term, write all the other ones in expanded forms can clearly see that the 5's and 7's Term of expanded form form an agp solve them but this becomes a bit calculative
the gen term is nth term of GP 7.10^(r+1) + summation till nth term of GP 5.10^r + 7 Then we just have to sum it all up, again there are some GPs and relate it with the given eqn.
Left JEE preparation 6 years ago and got this video in feed, managed to solve it after some time but I solved it from different method than you suggested. Yeah I felt proud solving that question. Also I can share the method. But I don’t think I can type it here.
Bhai maine to aise kia ki each preeceeding term mai 10 ka multiply karke 13 minus karde to next term mil jayegi. Isse maine general term nikali 77[10^(n-1)]-13{[10^(n-1)-1]}/9 Fir maine isse solve kiya aur ye answer aa gaya
I did it bhaiyya in 20 min by splitting terms 2 and 3 times I focused the most in maths during 11th (now going to 12th)at each and every topic that is helping me now to solve such problems
I solved it using general term method and it took me around an hour as i did so many calculation mistakes because of how underconfident i was that i can't solve a hard jee adv question, but i am glad i did solved it correctly at the end! At last i felt like it was just a 5-10 min question if done without errors, but koi na i will work on my calculation! Any tips on how to improve it?
Agar 11th mein ho to har ek sawal time bound karo nahi toh mere jaise L lag jayenge jee mein mai class mein topper tha maths mein bohot acha tha lekin jee ke time revision ke chakkar mein practice chut gya time bound uski vajah se ye haal hai
ALT APPROACH CLASS 5 TH METHOOD _____ WE CAN WRITE 77 = 7 .10@0 + 7 .10@1 here @ means power............ similarly we can write T(r) = 7 into 10 to power 0 + 5 into 10 @ 1+ 5 into 10@2 + 5 into 10@3 and so on.... + 7 into 10 @(r+1) . here this will GP and we can use sum of GP AND REWRITE GENRAL TERMS T(r)= some expression in ( r) and we can use sigma and it will get solved by doing some manipulation
@@sss-nd3ol terko yeh to pta hoga ki hm kisi bhi num for eg 77 ko 7 into 10 ki power zero + 7 into 10 ki power 1 is trh se rep kr skte hai SIMILARLY mai yha power ko * SE REP KRUNGA TO General term T(r) hai 7555........7 isko hm 7.10*0 + 5 .10*1 + 5.10*2 and so on 5 .10*r + 7.10 *(r+1) likh skte hai ab dekh yha 5 wali term main gp hai to formula use kr le or genral term aajaygi T(r) usme ab sigma aplly krnege to sum miljayega or hme que mein S= T99 +M /N NIKALNA AGAR DHYANN SE DKEHE TO TODI MANIPULATION KRNI PDEGI BUT HOJAYEGA
I have solved and my answer is 1292, let's check what's the answer Edit:I checked the solution and i have done silly mistake in calculation ans is 1219
After aadha page ki ghich-pitch I was able to think of this method and did it correct. I always thought my maths was weak but solving this really boosted my confidence. Now preparing for advance for 26th
The final S had 99 times 5 but you only multiplied by 10 then how it can be 75......57(99times 5) it only had75......57(98times 5). But if something I'm getting wrong please correct me with that.
Hii sir, maine ye sawal ke bare mai socha and abhi just Allen mai Sns ka test hua tho it kind of clicked ki gp hai difference. I accept final answer tak tho nahi aa payi par ¾ khud se figure ki. Thank you sir
I remember solving this question for IOQM last year in class 9th This was one of the only question i could do, and for me this was one of the easiest The problem is so many people get so absorbed in the advanced topics that they forget the basics of it all
I first found out the general term of 75...57 series and then calculated the value of S using finite sum of GP. The only time consuming part was to relate S with (75..57 +m)/n
I have not seen the solution (writing this comment after pausing the video). It is very easy to see that there are numbers with digit repeating which can be compactly represented as GP. There is yet another sum running over different GPs with different first term and last term. Kind of double summation which needs some manipulation to arrive at a compact representation using another GP sum which when reversed will give the repeating pattern number along with some constants m and n. I don't know if this is the solution described in the video (but I am taking a guess by just pausing and thinking about possible solution). It took me less than 30s to think all this. Does not seem like a hard problem at all.
By watching this series, i have learnt that almost none of the adv questions are difficult when solved with calm mind at home Solved this one also but this time it got a bit lengthy and spent approx 10-15 min💀 for the answer to come in the first attempt
the first stuff that clicked in my mind was ki pattern based hi sum hai...then deemag me aaya ki add subtract multiply krke kuch to hoga s ke sath n un dono ko minus krdenge jisse ki hamara denominator ayega n wala but upar wala smjh nai aa rha tha...pehele 1st 2nd and 2nd 3rd ko multiply kia dekhne ke lie kuch aage wale terms to nai ban rhe...nai bana...fir socha 10 se multiply krdu? then minus krne gye to -77 and 13 dekhkr laga nai..repeat nai ho rha hai ..age nai kia and tha ki last me ajayega so usko kaise hataungi...badme minus wale isse gp ka aaya..usse ban gya ...lekin was not sure..to numbers ko divide krke check kia 9.8 pehele ka then 9.83 smth aaya tb lga answer dekhna chahie...bht khushi mili ye dekhkr ki dono methods sai the😭 edit: sb no ko expand krne pr kuch patterns banenge? for eg 75557 ko 70000+5000+500+50+7??
This is actually not that difficult. We use similar method to derive the sum of GP. Suppose S = a + a.r + a.r^2 +...+a.r*n. The standard way is to multiply S with r and subtract the two series to get (r-1)S = a.(r^(n+1)-1). The important thing to realize is what should r be so that the terms either cancel or give a constant value.
Aah, the classic, the timeless, 10Sum minus 1Sum trick. Truly one of the staples and favourites of Sequence and Series questions. It is one of the most common, important, and crucial methods used in higher level Series questions. Thats why doing it was my first intuition as well.
The equation is 70(10^99 -1)/9 + 99x7 +50x(10T -98)/ 9 = ((70x10^99 +7 + 50( 10T+1) + m)/n Here I assume n is 9 and proceed further to eliminate T and we get m as 1210.
The technique you used is usually applied in Mathematical Induction. If someone has done a lot of induction sums then this approach may become apparent.
I am in class 10 and somehow found the approach in around 1 minute. I used S=77+770-13*11+7700-13*111 and so on. writing s like this i was able to find the value in 4-5 steps and then using same observation on 755...99times7 i got the value of m and n :). I am still surprised i was able to solve it cuz I usually get stuck on jee adv. pyqs
bhaiya maine ek method socha tha iske liye 80-3 800-43 8000-443 jaise har term ko likh lein fir uske baad hame ye dikh raha hai 3 ,43,443 agp hai with first difference in agp ham general term ko a(r^n) +b = T maan sakte hai please help me get ahead with this method . will really boost my morale.
In hindesight its good to take this approach, but to start visualizing , begin by expanding number , automatically the AGP form will click in brain. Its a simple one though. Things with double summations over two variable would eat up the mind more
I m not able to solve all the adv questions myself but did it really boosted my confidence great question I justed multiped with 10 and shifted the sum than breaked the last term into given
It took me 17 minutes to solve this question.. and here's my approach: By observation, we can get this expression : S=77+770-13*1+7700-13*11+77000-13*111+............+770000...(98 times)-13*1111...(98 times) Thereby , S=(77(10^99-1)/9) - 13 (1+11+111+1111+...........+111 ..(98 times) S=(77(10^99-1)/9) -13/9(+1-1 10-1+100-1+1000-1+....10^98-1) S=(77(10^99-1)/9)-13/9((10^99-1)/9 -99) Since,((10^k)-1)/9=1111....(k times) S=(77*10^99-13*1111...(99 times))/9 + 1210/9 Since,75...57=77*10^99-13*1111...(99 times) As stated earlier Therefore S=75.....57+1210/9 Therefore m+n=1219
Maine t2 aur t3 number tak sum likh ke us format mei likha aur fir mujhe pattern dikh gaya , m was in ap with 13 ki difference and N constant rahega wid 9 😊😊
One of the easiest question in paper yeah not as easy as binomial and statistics one, but I think most people would have gone wrong in calculation part as this method wouldn't strike in exam time, and going through normal like karte jao karte jao then it goes in approx 1.5 page of jee scribble pad and number of 5 ka gdbdi hua hoga sabke sath maybe. But ofcourse an easy question. Sometimes number of correct response doesn't give idea about difficulty of question.
ek approach ye aayi thi - sabhi do consecutive terms ka diff nikalo usme se 68 common lelo gp aaegi uska sum easily aajaega. thus now we have S and uske baad easy hi h....
wasnt that hard maybe looks scary but if you spend a minute or two and have practiced sequence and series before you can see how to split into different series
bhaiiii khud se ho gaya, but i did not use the method you used, i could not make that observation, i used recursion and i got the relation S(n+1)-S(n) - 77 = 68(10+10^2+10^3+.....+10^(n+1)) and then using the recursive relation converged it to S0 and then got the answer, took around 10 mins
I have solved this in 3rd try due to calculation mistake. In my method, I have expanded the number in powers of 10, uske baad to sawaal kuch bacha hi nahi.
Iss level ke sawal sachin sir class mein karwate the issi method se par main mand buddhi wahan bhi aur yahan bhi apni buddhi lagane par pada hoon..aine aise kiya ki s=sigma 2 se 99 (Tr). Isme (Tr) is sigma 7(10^r-1) + sigma 5(10^r-2) + ........7sigma(10)😅
I had this question solved in jee advanced 23 but still I couldn't even qualify my maths cutoff got 6/8 🤡 baaki subjects ka hogya tha. My sequence and series was good I guess.
I did solve it last year during my jee advanced exam by writing everything in summation form and adding it out but it took a fuck ton of my time unfortunately (5 mins)🥲 , I wrote 7(5..r)7 = 7x10^(r+1) + 7 + 5 sigma(10^r) now the general term is easy to solve as the summation 10^r is a simple gp but now we have to take summation again on both sides essentially solving another gp which is another headache . To get to the final answer we need to convert everything back to 7555..r7 form which isn't that hard. Now that I'm seeing this solution it was actually pretty clever once you notice difference as 13👍
Bro, also, like i think u splited all of the numbers into tens, hundreds and etc places... Then there will be many summations that u need to solve in order for u to get the answer... Then could u tell how did u solve it?
Dude , i solved it rn , im going by same method, but i dont get it , how i can get Sn from Tn I got Tn of new series formed by S 77 + 757 + 7557 +.... (70+7 ) + 700+50+7 + .... 70+ 98(7) + 97(5) + 96(500) + ..... I got Tn = (n+1)5(10^(97-n)) How do i get Sn now?hmm
Mtlb apan terms ko Tn= 75......57 =75........ 5+2 , kuch aise likh lete hai aur uske baad Tn= 7×10^(n+1) +5[10°+10¹+10²+10³...... +10^(n) ] + 2 ,, bs S= sigma Tn normal se khud se ajaye aise hi manipulation hain T99= n S- m, mai leke ana hai,, and done
4 minute lage Aise hi kiya tha aur 1st attempt me Pehle video dekha toh laga general twrm nikal k kiya aapne Aur maine 10 se multiply is liye kiya tha kyoki thodi observation se bich me ek sum1+11+111+1111... Krke bana toh maine soch isme 9 se divide multiply krne se achha shuru me krke dekhu aur consecutive difference bhi 68 wala dikh rha tha
Bhai yeh sawal literally jab solve kiye tha bahut aasan laga tha kyuki 1st try me aise hi kiye tha abb jab video dekh rha hu pat chal rha uss paper ka hardest sawal tha
Bhai I did in about 30 minutes because I kept doubting the procedure but after having confidence and calmly doing it for second time (calculation mistake in 1st time I got m>3000😅) I did it.
@@cycloneps4718bhai tension nahi lene ka. aap kya karo. physics aur chemistry ko 40 min se zyada mat do. try Karo 1.5ghante tak maths karne ka. chemistry m Kam score karoge chalega. but maths m 70+/80+ aana chahiye. pura Jaan Laga do. aur har question pagal ke tarah solve mat karo jee m. koi nahi puchega ki tumne kese solve Kiya. learn to skip steps in your mind. save as much time as possible. learn to be fast. 30 min to solve any question is a lot. I never spent more than 10 min for any question, even while practice.
How I did this question was : S = 77 + 757 + 7557 + ... + 75...57 S-99(7) = 70 + 750 +7550 +.....+75...50 S-99(7)(1-10)= 70 + 50 + 50 +...- 75...99...50 S= 755...99....57+99(13)-77/9 M= 99(13)-77 N=9 M+N= 1219 Solved in 5 mins because it took 3 mins to think how to go about this question .... Overall I would give this 5/10 rating
I think ye 5th class ki expanded form wali method se bhi ho jata let's see S= 70+7 + 700+50+7 + 7000+500+50+7 +______+ 7×10^99 + 5×10^98+(___)+7 then next step we simplify 7^99 + 5(1+11+111+1111+_____+11.....111) + 7(10+100+1000+____+100000.....) uske baad wahi 11th class ki general approach thoda calculation hogi.... but ansewer aa jayega BTW apka method jyada easy h 👍
Bhaiya , I need to ask one thing that even if answers humare aa rhe hain , it takes a lot of time... like 14 - 15 mins lg rhe solve krne mein... but advanced ke paper mein itna time lga denge fir toh mushkil ho jaayegi.. time required to solve the question (**that too with alertness**) kese kam kre?
General term wale method se try kiya..phir kuch calculation mistakes kardi aur phir aage phass gaya.... anyways second solution looks elegant but critical thinking required
Bhaiya mene ise dusre method se kara Mene pehle 7555...57 ki general value nikali gp sum ke formule se, fir mene us expression ko summation mein rakha. Fir gp ke sum ke formule se use calculate kara. Fir jese expression pehle ban rhi thi use mene convert jarne ki koshish kari aur fir answer aa gaya Ye mene video dekhe bina kara hai Edit after watching video: bhaiya ka solution jaada easy tha. Par mujhe mera solution karne mein maza aaya
Sachin sir ne ye hi method bataya tha esi series ko solve karne ke liye is that's why solution dekhne ke bad kuch khas nhi laga but me ise exam me paka chod deta because mere se series hi nhi banti
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i solved it in 2 days🙂
I did it this way:
first found out a general term for S by observation as: 7*10^r+1 + 5*10^r-1+....+ 5*10+7 for a general 'r'. then just apply summation of this from r=0 to 98 and use gp formula on 10^r+1 and compare with the other form to get m and n.
this is the most basic approach that can be thought of during exam i think, that 10S one is too intuitive for me.
same, thats what I did as well
same
instead use general form 7x(10^n) + 5/9{(10^n) - 1} + 2
Same
@@Realsxullzsame bhai Maine bhi aise hi Kiya tough utna nahi tha isse 6 min me hogaya
10s wala method ko apply karne ke liye 13 ka table acha se yad hona cahiye tab hi wo click hoga ki Haan aise bhi kar sakte hain
Bhaiyon, 85 batch IITK. Kiya tha yeh sab ek zamane mein. Aisa laga ki kabhi kaam na aya per aise sawalon se dimag tez ho jaata hai - kahin bhi kisi se bhi compete kar paoge! Good luck to you all! IIT mein milenge!
What I understood from the question is that when the terms of the series are increasing digit by digit you should multiply by 10^(no of digit increases per term)
As a guy who cracked Neet.. good thing i didnt take maths 💀. 1000 page ki books ratt loonga but yeh mere bas ke bahar hai😂😂
Bro can you tell some effective methods for remembering the text?
@@arihantsingh5953 skill issue
Well to each their own. Mujhe utna rattna pasand nhi 😂 pehle hum soche the neet de denge but accha hua maine biology time rehte chor diya 💀 11-12 ka bio mere bas ke bahaar hya; isse accha maths kar lunga, accha bhi lagta hya maths
*"Bruh why all the people in the comments are telling this was easy and even some are telling that they solved it orally (sarcasm)💀☠️. Bro ur whole comment is filled with toppers having AIR
Un sb ki m k c
😂😂
aree ye sab fekte bohot hai. asli paper ka reality yahi h 50% marks bhi score nahi kar pate 99.8% of jee advanced aspirant
that's a fact you cannot deny
par after solving the question, it was actually really mains level question ngl.
maybe exam pressure hoga baccho pe
@@godson200 aa gye aukat pe , itna pravachan dene ke baad
I split the term and got some value, pretty sure its correct but getting it in terms of the question was hard, nice method bro
thank you bhaiya...doing every lecture
good going bro.
Bhaiya I swear diwali ki chutti me yehi sawaal kiya tha 1st try me ho gya muskil nhi laga tha ☠️☠️
Bhai ye Pathfinder ka sawal hai mai karta tha fun mein bachpan mein uska solved exam hai usme method tha 10 se multiply krdo aur subtracr krdo fir gp ban jayega 2 min mein ho gya mujhe fix approach pata tha aisa mere saath kash mere paper mein bhi ho jaye that's why reading solved example are a must
Bhai mei aaise Salwa 6th class se karta aaraha hu
@@piyushjain7687 average jain 🛐🛐🛐 bhai jain aur agarwal kis mitti ke hote hai I can trust anyone easily mai toh tomar hu naam bhi nahi ata kisi ka top 1k mein bas achi baat ye hai I'm obc
@@Ayush-mg6xw majak kar Raha hu Mari preparation mei bss physics ke L Lage hai 😣 pathfinder try karne ke liye base hi nhi bann pa Raha hu maths badiya hai 🥵🗿bss iss questions ko try Kiya tha first attempt meaning bana tha because GP padhne ke direct baad iss question pe kud gaya tha
@@piyushjain7687 bhai tu jain hai under 500 pakka
It was quite easy , the method was simple , the catch is multiplying S by suitable number to get the answer
Bro but us time par ye sochna bhi to hard hai , ye waise board level par bhi ata hai question
@@YamanDiwan yeah you got that right . But by practicing more and more and staying calm really helps you think out of the box
@@Notsosarcastic_02 so you are in 11th right
@@YamanDiwan i am a dropper 😂 scored 96 in jan
I was able to solve it in under 10 minutes using a much simpler and obvious approach. You can try adding up the unit, tens up till ten thousandth digit by adding 7*99 + (5*98 + 7)*10 + (5*97 + 7)*100 and you get 61683 as the last five digits. You know the sum must be of at least 100 digits so remainder “n” can not be greater than 12 otherwise the numerator would be greater than 101 digits which shouldn’t be possible since m and n are less than 3000. Then you can multiply the last 4 or 5 digits with each number up to 12 and will easily be able to note when you get to 9 that it’s the only time the multiple is greater than the sum in the threshold of 3000. You can continue to check a few more digits when multiplying by 9 but will see it easily and see the multiple ending in 6767 which is 1210 more than 5557 and you get 1219.
Too lengthy
@@dakshsingh5891 True, it do be quite lengthy but simple enough to think in a minute and can still be done in 5 minutes
@@tilakagarwal8163yeah bro.. I solved it in the same exact way itself..it's not that lengthy though
@@tilakagarwal8163still lengthy. Rather than multiple S by 10 then subtract Then S will become 9S=13*98-77+755555....570
@@ManasChoudhary-p1xbruh at that point you’re just guessing the remainder to be 9 and hoping to be correct, it’s like saying why don’t you hit in the first trial of hit and trial
Another way of solving this question is by using method of difference which a lot of people don't really know about.
If you notice, you'll find out that the common difference between consecutive terms is in GP. Therefore by using method of difference the general term can be given by a(r)^n-1 + b where 'a' and 'b' are variables, r is the common ratio of gp. Using n=1 ans n=2 you can find the value of a and b and thus you'll get the general term. Now just use sigma on general term and do some basic manipulation to get the answer.
This is how I solved it during my first attempt.
Edit - didn't see bhaiya already mentioned it in the video. I just commented without seeing the video
To delete maar abhi.. :)
@@SANDRA-xq7es 😢
@@rawatutkarshBhai fir bhi sirf 424 bacche solve kiye . I mean ye out of the box ques bhi nhi tha only bhaiya ka method was like too out of the box . Actually me exam situation kuch level of difficulty badha deta hai or else Ghar me baithke toh kitne saari ne solve kar diya
@@vyukastra sahi bol raha hai bhai
Coaching tests me bhi yahi hota hai simple sawal time pressure me mushkil lagne lagte hai
One other method is that - notice that difference of succesive term is in gp (First term 680; r=10) we can simply apply Tñ=arñ+b
I think most simplest method
I did with that but it give the result tn-t1 = sum of that gp
How to proceed further
That's how I did it too
Method of difference when common difference is in gp 🙇
@@piyushsharma18941 find a and b tn = ar^n+b then find tn then make Sigma and solve it
I did break
77 = 22+55
757 = 202 + 555
7557 = 2002 + 5555
Now 2002 = 2(10^3 + 1)
5555 = (5/9)(10^4-1)
And that's how I solved it, but it was still calculation expensive.
Just takes around 5 minutes to figure out n=9 using either
- Principle of Recursion
- Asymptotic analysis
- Linear algebra ( eigen analysis)
konsi class me ho ?
chod liya gyan ban gya hero?
@@utkarsh3945 It's a basic question testing intuition, Taught very early in advanced sequences and series.
& for m
We need to solve collatz conjecture😂😂
How did u do it by linear algebra.. Can u explqin more
Sameer bansal definite integral Ex 5 ke Q.50 me ek result tha ki (111...11) n times= 10^(n) -1/9 waha se sochke pattern bana liya (755...57)=(777..77) (n+2) times - 10 x (222..22) n times, iske baad bas calculation aur consolidate karna tha.
Yup
aree bhai bhai 🙌🏻
sir mene i solved this today morning what i did was write 77 as 70+7 , 757 as 700+50+7 , 7557 as 7000 +500+50+7 ....
which we will become s = 7*99 + 7*(10+100+1000....) + 50( 1+ 11 +111...)
phir isko solve kara aur ansar agaya roughly 8min lage the
Same mene bhi Yahi socha tha
pr mene aage ki calculation nhi ki
kyuki mujhe maloom tha
ki ye general approach h to answer aa hi jayega
BTW well done !!❤
I solved it with a completely different approch
jese bachpan me likhe the wese
77
+ 757
+7557........ likha
then
isme pattern notice kiya ki is likh sakte hai as
7(99) [ones place ke sab 7 ka sum]
+7(10+100+1000.....10^99) [Remaining 7s according to its place]
+5(10+100+1000.....10^98) [Remaining 7s ke just right wale 5s]
+5(10+100+1000.....10^97) [just right wale aur 5s]
.
.
.
5(10) [aakheri tens place wala 5 jo series ka last digit hai]
then
(10+100+......) wale part me GP ke sum ka formula lagaya
75...99.....57 wala term dikh gya +57 -57 karna para kyuki initially
75............500 form hua tha
then answer 1219 aa gya
{First attempt me
starting me
7(100)+......... [100 nhi 99 hona tha]
kar diya tha
ans 1282 aaya tha fir video dekha toh pata chala
ans 1219 hai thora soochne ke baad galti dikh gai ki
7(99)
+7(10+100+.......) hoga
then correct answer aa gya }
hamare sir ne same method se karaya nice bhai
Chutiya tarika hai yah 😂 bas calculation karta jao jao orh time waste hai orh kuch nhi . Rather than Asa karo S=77+757+7557+. N
Multiple by 10 on both the side 10S=770+7570+75570+755570. 755..............570 Now subtract both then you will observe a pattern 9S=-77+13+13+13+13+13+13. N +7555555555.........70
Now 9S=13*98-77+7555555.......570
S=13*98-77+755555555.....70/9
Hence M+n is 1219@@4fgaming925
nice approach
Yeh taarika PnC mein sikhaya jata hai
Bro I tried same method but my sequence bit weak so left it after writing it all in summation format.
Bhaiya ek time tha jab aapke videos dekh ke dar lagta tha, ab to comments padhke aur bhi jyada dar lagne laga hai
😂😂
I was the only one in my coaching to have solved question like this alongside my teacher (obviously), in questions like these where some kind of recurrence pattern is forming, this AGP method works usually, a few terms will first trouble you but with enough practice it becomes easy
So trust yourself and work hard
Can be done by writing expansion don't take that 99th term, write all the other ones in expanded forms can clearly see that the 5's and 7's Term of expanded form form an agp solve them but this becomes a bit calculative
ohh, I see 👀
This shows lack of practice dude it is probably the worst method to solve such kinds of series
the gen term is nth term of GP 7.10^(r+1) + summation till nth term of GP 5.10^r + 7
Then we just have to sum it all up, again there are some GPs and relate it with the given eqn.
Left JEE preparation 6 years ago and got this video in feed, managed to solve it after some time but I solved it from different method than you suggested. Yeah I felt proud solving that question. Also I can share the method. But I don’t think I can type it here.
Bhai maine to aise kia ki each preeceeding term mai 10 ka multiply karke 13 minus karde to next term mil jayegi. Isse maine general term nikali
77[10^(n-1)]-13{[10^(n-1)-1]}/9
Fir maine isse solve kiya aur ye answer aa gaya
Apne answer ko please thoda elaborate karna ki aapki general term kaise bani?
Use recurrence relation ... More smart way of solution....
I did it bhaiyya in 20 min by splitting terms 2 and 3 times
I focused the most in maths during 11th (now going to 12th)at each and every topic that is helping me now to solve such problems
Mam koi tips for juniors entering into 11th?
@@djshadowpo2.023 be consistent either study 3hrs or 8hrs but study daily and revise everything u study on that day only
I solved it using general term method and it took me around an hour as i did so many calculation mistakes because of how underconfident i was that i can't solve a hard jee adv question, but i am glad i did solved it correctly at the end! At last i felt like it was just a 5-10 min question if done without errors, but koi na i will work on my calculation! Any tips on how to improve it?
What was the general term that u got?, was it like summation of (r+2) from 0 to 99?
Agar 11th mein ho to har ek sawal time bound karo nahi toh mere jaise L lag jayenge jee mein mai class mein topper tha maths mein bohot acha tha lekin jee ke time revision ke chakkar mein practice chut gya time bound uski vajah se ye haal hai
ALT APPROACH CLASS 5 TH METHOOD _____ WE CAN WRITE 77 = 7 .10@0 + 7 .10@1 here @ means power............
similarly we can write T(r) = 7 into 10 to power 0 + 5 into 10 @ 1+ 5 into 10@2 + 5 into 10@3 and so on.... + 7 into 10 @(r+1) .
here this will GP and we can use sum of GP AND REWRITE GENRAL TERMS
T(r)= some expression in ( r) and we can use sigma and it will get solved by doing some manipulation
can u pls explain this method again? thank you so much!
@@sss-nd3ol terko yeh to pta hoga ki hm kisi bhi num for eg 77 ko 7 into 10 ki power zero + 7 into 10 ki power 1 is trh se rep kr skte hai
SIMILARLY mai yha power ko * SE REP KRUNGA TO
General term T(r) hai 7555........7
isko hm 7.10*0 + 5 .10*1 + 5.10*2 and so on 5 .10*r + 7.10 *(r+1) likh skte hai
ab dekh yha 5 wali term main gp hai to formula use kr le or genral term aajaygi T(r)
usme ab sigma aplly krnege to sum miljayega or hme que mein S= T99 +M /N NIKALNA AGAR DHYANN SE DKEHE TO
TODI MANIPULATION KRNI PDEGI BUT HOJAYEGA
hmm, not bad.
I did same :D
I have solved and my answer is 1292, let's check what's the answer
Edit:I checked the solution and i have done silly mistake in calculation ans is 1219
wow topper lmao
Bhaiyaa, kattai jeher sawal hai... Karke maza aa gaya... 😎😎
After aadha page ki ghich-pitch I was able to think of this method and did it correct. I always thought my maths was weak but solving this really boosted my confidence. Now preparing for advance for 26th
The final S had 99 times 5 but you only multiplied by 10 then how it can be 75......57(99times 5) it only had75......57(98times 5). But if something I'm getting wrong please correct me with that.
After multiplying with 10 the last digits became 570, then he split it as 557+13 , where we get another 5. And got a number with 99 5's.
Check at 9:26
@@parjanyak1104👍
@@parjanyak1104 got it.
Hii sir, maine ye sawal ke bare mai socha and abhi just Allen mai Sns ka test hua tho it kind of clicked ki gp hai difference. I accept final answer tak tho nahi aa payi par ¾ khud se figure ki. Thank you sir
I remember solving this question for IOQM last year in class 9th
This was one of the only question i could do, and for me this was one of the easiest
The problem is so many people get so absorbed in the advanced topics that they forget the basics of it all
So true
Bhai 10th me hu under 10mins me solve hogya aur bilkul same solution tha mera or apka ❤
Hlo
Ok... tum hi laoge AIR 1 JEE 2027 Mai.
@@a.m.videos3254 bro 2026 mein
@blaareen. Maine assume Kiya ki uska session just start Hua
11th me hojaonha iss saal❤@@a.m.videos3254
2nd method is just AGP , used as further generalization of GP
Yeh method se to mene bohot easily Solve karliya.
I first found out the general term of 75...57 series and then calculated the value of S using finite sum of GP. The only time consuming part was to relate S with (75..57 +m)/n
I have not seen the solution (writing this comment after pausing the video). It is very easy to see that there are numbers with digit repeating which can be compactly represented as GP. There is yet another sum running over different GPs with different first term and last term. Kind of double summation which needs some manipulation to arrive at a compact representation using another GP sum which when reversed will give the repeating pattern number along with some constants m and n. I don't know if this is the solution described in the video (but I am taking a guess by just pausing and thinking about possible solution). It took me less than 30s to think all this. Does not seem like a hard problem at all.
By watching this series, i have learnt that almost none of the adv questions are difficult when solved with calm mind at home
Solved this one also but this time it got a bit lengthy and spent approx 10-15 min💀 for the answer to come in the first attempt
"solved with calm mind at home" yeah they dont give you that at the exam centre. Jee is hard for a reason
@@prismatic-bl8qf wahi baat mein point out karna chah raha tha
Examination temperament and confidence on preparation are very important
the first stuff that clicked in my mind was ki pattern based hi sum hai...then deemag me aaya ki add subtract multiply krke kuch to hoga s ke sath n un dono ko minus krdenge jisse ki hamara denominator ayega n wala but upar wala smjh nai aa rha tha...pehele 1st 2nd and 2nd 3rd ko multiply kia dekhne ke lie kuch aage wale terms to nai ban rhe...nai bana...fir socha 10 se multiply krdu? then minus krne gye to -77 and 13 dekhkr laga nai..repeat nai ho rha hai ..age nai kia and tha ki last me ajayega so usko kaise hataungi...badme minus wale isse gp ka aaya..usse ban gya ...lekin was not sure..to numbers ko divide krke check kia 9.8 pehele ka then 9.83 smth aaya tb lga answer dekhna chahie...bht khushi mili ye dekhkr ki dono methods sai the😭
edit: sb no ko expand krne pr kuch patterns banenge? for eg 75557 ko 70000+5000+500+50+7??
This is actually not that difficult. We use similar method to derive the sum of GP. Suppose S = a + a.r + a.r^2 +...+a.r*n. The standard way is to multiply S with r and subtract the two series to get (r-1)S = a.(r^(n+1)-1). The important thing to realize is what should r be so that the terms either cancel or give a constant value.
Aah, the classic, the timeless, 10Sum minus 1Sum trick. Truly one of the staples and favourites of Sequence and Series questions.
It is one of the most common, important, and crucial methods used in higher level Series questions. Thats why doing it was my first intuition as well.
u a jee aspirant?
Same here my intuition was same as well but i didn't solve it lol!
The equation is
70(10^99 -1)/9 + 99x7 +50x(10T -98)/ 9 = ((70x10^99 +7 + 50( 10T+1) + m)/n
Here I assume n is 9 and proceed further to eliminate T and we get m as 1210.
Cengage ke illustration me tha same ye question
kuch bhi lol main puri cengage kiya hu aaisa qn hi nhi hai wala
The technique you used is usually applied in Mathematical Induction. If someone has done a lot of induction sums then this approach may become apparent.
I am in class 10 and somehow found the approach in around 1 minute. I used S=77+770-13*11+7700-13*111 and so on. writing s like this i was able to find the value in 4-5 steps and then using same observation on 755...99times7 i got the value of m and n :). I am still surprised i was able to solve it cuz I usually get stuck on jee adv. pyqs
bhaiya maine ek method socha tha iske liye 80-3 800-43 8000-443 jaise har term ko likh lein fir uske baad hame ye dikh raha hai 3 ,43,443 agp hai with first difference in agp ham general term ko a(r^n) +b = T maan sakte hai please help me get ahead with this method . will really boost my morale.
group join karlo,
@@jeesimplified-subject bhaiya link mil sakta hai kya please 🥺
Solved in 2.5 mins (inclusive of reading time)
Air 0 ho ka?
In hindesight its good to take this approach, but to start visualizing , begin by expanding number , automatically the AGP form will click in brain. Its a simple one though. Things with double summations over two variable would eat up the mind more
I m not able to solve all the adv questions myself but did it really boosted my confidence great question I justed multiped with 10 and shifted the sum than breaked the last term into given
It took me 17 minutes to solve this question.. and here's my approach:
By observation,
we can get this expression :
S=77+770-13*1+7700-13*11+77000-13*111+............+770000...(98 times)-13*1111...(98 times)
Thereby ,
S=(77(10^99-1)/9) - 13 (1+11+111+1111+...........+111 ..(98 times)
S=(77(10^99-1)/9) -13/9(+1-1 10-1+100-1+1000-1+....10^98-1)
S=(77(10^99-1)/9)-13/9((10^99-1)/9 -99)
Since,((10^k)-1)/9=1111....(k times)
S=(77*10^99-13*1111...(99 times))/9 + 1210/9
Since,75...57=77*10^99-13*1111...(99 times) As stated earlier
Therefore S=75.....57+1210/9
Therefore m+n=1219
I solved it by creating a general term for 75...57,
T =75(...n times)7 = (13 + 680 * 10^n)/9
if n = 0 => T =77,
if n = 1 => T = 757
Maine t2 aur t3 number tak sum likh ke us format mei likha aur fir mujhe pattern dikh gaya , m was in ap with 13 ki difference and N constant rahega wid 9 😊😊
thank you bhaiya at the beggining i paused and tried it, it is an easy question unless you dont know about agp
One of the easiest question in paper yeah not as easy as binomial and statistics one, but I think most people would have gone wrong in calculation part as this method wouldn't strike in exam time, and going through normal like karte jao karte jao then it goes in approx 1.5 page of jee scribble pad and number of 5 ka gdbdi hua hoga sabke sath maybe.
But ofcourse an easy question. Sometimes number of correct response doesn't give idea about difficulty of question.
Bhaiiya easy laga tha....practice itni karli hai sequence ki...Sum dekh kr hi idea laga tha ki T2-T1 wala method lagega
Man, I solved this problem in the exam centre itself (YAY!) and got it right.
Bhaiya ye method AGP solve krne ke method jesa hai bas direct nhi hai difference gp bana ra tha ho gaya solve mujhe bhi ye
Thanku
ek approach ye aayi thi -
sabhi do consecutive terms ka diff nikalo usme se 68 common lelo gp aaegi uska sum easily aajaega. thus now we have S and uske baad easy hi h....
Bhaiya question bahu accha laga aur jis tarah apna explain kiya question easy ho gaya
Question Easy Tha But Mere Solution, Ne Hi Mujhe Uljha Diya
wasnt that hard maybe looks scary but if you spend a minute or two and have practiced sequence and series before you can see how to split into different series
bhaiiii khud se ho gaya, but i did not use the method you used, i could not make that observation, i used recursion and i got the relation S(n+1)-S(n) - 77 = 68(10+10^2+10^3+.....+10^(n+1)) and then using the recursive relation converged it to S0 and then got the answer, took around 10 mins
Math is beautiful!
I did this but with a complete different approach like by replacing each term of S with 99 repetition wali terms
I have solved this in 3rd try due to calculation mistake. In my method, I have expanded the number in powers of 10, uske baad to sawaal kuch bacha hi nahi.
Aj dekhne Mai thoda late ho gya boards ka kr rhi thii🥱😂btw I solved the question half✨🔥
i did it by the longest and most calculative method possible , took me 25 minutes , i just kept on solving and was about to leave it
My approach:-
77=20+55+2
757=200+555+2
7557=2000+5555+2
And this goes on...
Ab we know how to calculate the sum of these 3 series
Bhaiya cengage ka ek sawal isse match karta hai... Exact nhi hai lekin us question ka concept use hua...
hi can you tell me how you get these stitistics of how many people attempted and and correct
search for jee advance report, you will a huge pdf, go to question paper analysis in that
Iss level ke sawal sachin sir class mein karwate the issi method se par main mand buddhi wahan bhi aur yahan bhi apni buddhi lagane par pada hoon..aine aise kiya ki s=sigma 2 se 99 (Tr). Isme (Tr) is sigma 7(10^r-1) + sigma 5(10^r-2) + ........7sigma(10)😅
I had this question solved in jee advanced 23 but still I couldn't even qualify my maths cutoff got 6/8 🤡 baaki subjects ka hogya tha. My sequence and series was good I guess.
same aisa swal kr chuka hum
prove that n(4s)n-1(8s)9 always form q perfect square
I did solve it last year during my jee advanced exam by writing everything in summation form and adding it out but it took a fuck ton of my time unfortunately (5 mins)🥲 , I wrote 7(5..r)7 = 7x10^(r+1) + 7 + 5 sigma(10^r) now the general term is easy to solve as the summation 10^r is a simple gp but now we have to take summation again on both sides essentially solving another gp which is another headache . To get to the final answer we need to convert everything back to 7555..r7 form which isn't that hard. Now that I'm seeing this solution it was actually pretty clever once you notice difference as 13👍
Bhai tumhare mains me kitni percentile bani... U seem to be quite good in maths
11th me ho ki 12th me?
Bro, also, like i think u splited all of the numbers into tens, hundreds and etc places... Then there will be many summations that u need to solve in order for u to get the answer... Then could u tell how did u solve it?
Dude , i solved it rn , im going by same method, but i dont get it , how i can get Sn from Tn
I got Tn of new series formed by S
77 + 757 + 7557 +....
(70+7 ) + 700+50+7 + ....
70+ 98(7) + 97(5) + 96(500) + .....
I got Tn = (n+1)5(10^(97-n))
How do i get Sn now?hmm
Mtlb apan terms ko
Tn= 75......57 =75........ 5+2 , kuch aise likh lete hai aur uske baad
Tn= 7×10^(n+1) +5[10°+10¹+10²+10³...... +10^(n) ] + 2 ,, bs S= sigma Tn normal se khud se ajaye aise hi manipulation hain
T99= n S- m, mai leke ana hai,, and done
4 minute lage
Aise hi kiya tha aur 1st attempt me
Pehle video dekha toh laga general twrm nikal k kiya aapne
Aur maine 10 se multiply is liye kiya tha kyoki thodi observation se bich me ek sum1+11+111+1111... Krke bana toh maine soch isme 9 se divide multiply krne se achha shuru me krke dekhu aur consecutive difference bhi 68 wala dikh rha tha
Bhaiya maine toh general twrm nikal li
7. 10^r +5/9 .(10^(r-1) -10) +7
Or puree pe sigma laga diya gp aue ap mai tooot gaya question
I tried and got it right in 20 minutes, but in my case i need to do some manipulations in the given sum as well which took me a time
20 minutes for this??
thought of multiplying the sum by 10 and evaluating but now i regret seeing the solution
great solution🙌
Bhai yeh sawal literally jab solve kiye tha bahut aasan laga tha kyuki 1st try me aise hi kiye tha abb jab video dekh rha hu pat chal rha uss paper ka hardest sawal tha
How much interested this question but I don't solve this , directly see the solution but understand everything ☺️
Question asaan hai but needs a little bit of patience and calmness !!! ❤
Bhai I did in about 30 minutes because I kept doubting the procedure but after having confidence and calmly doing it for second time (calculation mistake in 1st time I got m>3000😅) I did it.
You tried it for so long, I appreciate your passion.
Thank you, confidence stepped down after mains result( maths=33 marks)
@@cycloneps4718bhai tension nahi lene ka.
aap kya karo. physics aur chemistry ko 40 min se zyada mat do. try Karo 1.5ghante tak maths karne ka. chemistry m Kam score karoge chalega. but maths m 70+/80+ aana chahiye. pura Jaan Laga do.
aur har question pagal ke tarah solve mat karo jee m. koi nahi puchega ki tumne kese solve Kiya. learn to skip steps in your mind. save as much time as possible. learn to be fast. 30 min to solve any question is a lot. I never spent more than 10 min for any question, even while practice.
Solved this question within 11 minutes....tbh it was similar to cengage examples question ,if one has practiced them thoroughly
used almost the same method just simplified it using GP
bhai tum kis coachichng se kitna sal prepare kar rahe ho bhai?
@@JiTendra-m3t Self Prepration 💀
Bhaiya mujhe ye question 4th try me ho gya I am im 11th. 😊😊
I solved it in 3 minutes! I am from IIT Roorkee, now living in US
@@666Lindane I am pursuing PhD in Aerospace engineering
merse solve hogya sir. bhot maja aaya solve krke
How I did this question was :
S = 77 + 757 + 7557 + ... + 75...57
S-99(7) = 70 + 750 +7550 +.....+75...50
S-99(7)(1-10)= 70 + 50 + 50 +...- 75...99...50
S= 755...99....57+99(13)-77/9
M= 99(13)-77
N=9
M+N= 1219
Solved in 5 mins because it took 3 mins to think how to go about this question .... Overall I would give this 5/10 rating
Bhaiya what a great solution
10 T(r) - 13 = t(r+1)
9T(r) - 13 = t(r+1) - t(r)
summing and getting ans
ab soln bhi jldi jldi dekhleta
I am in 10th but yeh I was able to do upto some limit..😅
As I was preparing for olympiad level questions..
I think ye 5th class ki expanded form wali method se bhi ho jata
let's see
S= 70+7 + 700+50+7 + 7000+500+50+7 +______+ 7×10^99 + 5×10^98+(___)+7
then next step we simplify
7^99 + 5(1+11+111+1111+_____+11.....111) + 7(10+100+1000+____+100000.....)
uske baad wahi 11th class ki general approach
thoda calculation hogi....
but ansewer aa jayega
BTW apka method jyada easy h 👍
bro time limite hota hai
Bhaiya , I need to ask one thing that even if answers humare aa rhe hain , it takes a lot of time... like 14 - 15 mins lg rhe solve krne mein... but advanced ke paper mein itna time lga denge fir toh mushkil ho jaayegi.. time required to solve the question (**that too with alertness**) kese kam kre?
General term wale method se try kiya..phir kuch calculation mistakes kardi aur phir aage phass gaya.... anyways second solution looks elegant but critical thinking required
Intution lagana tha , thoda calculative tha , lekin hogya easily in the sense without any exam pressure
Bhaiya mene ise dusre method se kara
Mene pehle 7555...57 ki general value nikali gp sum ke formule se, fir mene us expression ko summation mein rakha. Fir gp ke sum ke formule se use calculate kara. Fir jese expression pehle ban rhi thi use mene convert jarne ki koshish kari aur fir answer aa gaya
Ye mene video dekhe bina kara hai
Edit after watching video: bhaiya ka solution jaada easy tha. Par mujhe mera solution karne mein maza aaya
Perfect ⚡️
Sachin sir ne ye hi method bataya tha esi series ko solve karne ke liye is that's why solution dekhne ke bad kuch khas nhi laga but me ise exam me paka chod deta because mere se series hi nhi banti