Kitni Bechain Hoke Ft DIVINE x MC STAN | Dip SR | DRILL MASHUP
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- Опубліковано 28 вер 2024
- ♫ Kitni Bechain Hoke Ft DIVINE x MC STAN
♫ Mashup By : Dip SR
♫ Visual : Jayant Visual
♫ Thumbnail By : Sukhen Visual
Uploaded for promotional and preview purposes only!
If you as a copyright holder wish to remove this,
please contact me : - tanmoysarkarofficial0@gmail.com
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► Used Tracks Credits :
♫ Kitni Bechain Hoke :- • Kitni Bechain Hoke | U...
Movie: Kasoor
Singer : Udit Narayan & Alka Yagnik
Music : Nadeem-Shravan
Lyrics: Sameer
Label : Saregama India Limited
Year : 2001
♫ Satya :- • DIVINE - Satya | Prod....
Artist - DIVINE
Lyrics - DIVINE, Gulzar
Composed by - DIVINE, Vishal, Karan Kanchan
Produced by - Karan Kanchan
Original Sample - Goli Maar Bheje Mein (Venus Worldwide Entertainment Pvt. Ltd)
Mixing & Mastering - Hanish Taneja
Label - Mass Appeal India / Gully Gang
English Translation - Nabeel Lakhani
♫ MC STΔN - TADIPAAR :- • MC STΔN - TADIPAAR | O...
Singer / performer: MC Stan
Music Composer: MC Stan
Author / Lyrics: MC Stan
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#KitniBechainHokemashup #DIVINE #MCSTAN
happy valentines day guys
Fuck valentine's day pray for our soldiers 🇮🇳🇮🇳
Maze hi mazee 🌚
😢broooo
@@bubaisarkar888 0❤❤❤❤❤❤❤
😮😮😮😮😮😮❤❤❤❤❤
Divine हां 🔥🔥🔥 creat goosebumps only divine fans
Nice😊😊😊😊😊❤❤❤🎉🎉🎉🎉😅😅😅😢🎉😂😮😅😊🎉😂
Team work best hai bhai ❤
😢😢ç😢 0:55 0:55 🎉j
Ib😊
S,
Cigarette ke bajase khatam ho raha stamina😂
Next level Mashup❤🙏🙏🙏
😂😂😂😅
ओरिजिनल गाने की इज्जत लूट ली भाई
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😂
😂😂😂
105301
😂😂😂😂😂😂😂😂😂😂
Viral kno hoini eta akhn oooo!!,
Amazing execution dada ❤️
Thanks bhai 🥰🥰
I am from Bangladesh 🌹🥀🥰 one of the best Rap.....
Thanks bro
Hi
Deadly combination ❤
Amazing ❤❤❤
Machate raho 🔥🔥
80 hajar ke shoes hai 😎
Divine's voice awesome
Ye abhi mera favourite song hogya ❤❤
Daily 10/15 baar sunta hoo
1 baar ...Silla yeh chahat ka( .mc ) ka ....1 nambar song hai.❤
Silsila yeh chahat ka .
Ma kasam aisa gana nhi suna kabhi❤😊😊
Iam from Bangladesh but only divine fan bro frie
Superb mix up
Best rap 🔥🔥🤯🤯
good remix rap song,❤
Gjjb ❤❤❤❤ 0:40
Uff so nice song
Next - 90s Bollywood Mashup Bhaii Please ❤🙏 Purane saab acche acche songs dalke 🥺
Aschee
Mc stan ko dil se mante hai ❤❤❤
DIVINE AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
Nyc bro 👍
Thanks bro
Hii🥰🥰🥰
King
Awsome ❤❤
❤❤❤😊😊😊
Gobind editor Raj bhar ji 💪💪💪💪😈💪
🙌🏻😍👑
D I V I N E 💀
Mera bhai nast ❤❤❤
🔥🔥🔥🔥🔥🔥🔥🔥
❤❤❤❤
💓💓
❤❤🎉 love you bro
😅😅😅😅😅😅😅নাইস সং😅😅😅😅😅😅
❤❤❤❤❤❤
Divine.4.10 ❤❤❤❤❤❤❤
Divine only ❤
Gall se h na hun 😘😘 h na
🥵🥵🥵🥵🔥🔥
Nice song bhai
Stanny👁️🥵🔥
😊😊
😮😮😮😊😊😊
Mc❤❤❤
Divine is tha best
❤❤❤🎉🎉🎉🎉🎉🎉
Ashish kumar ❤❤❤❤❤❤
Sabbash Dip$R
Thanks
Key bro mix acha hai prntu mai for solo version ke liye aya tha. But mix acha hai lakin because of i disappointed ulike. Please make it solo.
UA-cam Recomend Me Listen To Best Song Ever❤❤ Love Dip Sr 😍
____________________________My Love_____________________________My Heart
How ❤😊
Accha hua kitna bechen hoke gana h mix mai nhi to mai sun hi nhi pata
😈👹💂♂🙏👑💪🙏🌹🙏🥃🥃🥃🥃🥃🥃गोलुरवत
Hi
💔💔💔💓💗💓💓
ashish kumar love two priyanka kumre ashish4kumar 750
Legend x Chhapri
❤❤❤❤
❤❤❤❤❤❤❤
0000000
MC STAN_❤️
DIVINE❤
mazaK mazaK M masteR piecE bnA detA...😅😅😅
Divine bhai to aag laga diye
Andhere me chamkunga nahin dekhna mujhe savera so deep line bro ❤
Ashishkumar pawn Priyanka ❤❤❤❤
Divine ft kitna becain hoke this fire song 🎉
Kitni bacha mn ho k nice song 🎵 but I’m fine ❤
Jab me car me music chLata hm TB papa ko Jo me lagata tha pasand Nahi aata tha pr jab ye song car me play karta hm to papa kuch bol Nahi pate 😁😂😀
Very Nice Love Song 💯
Thanks bro
nice singh❤❤
Nice mc altaf baba go high boy god blessed u iam so happy ❤️❤️❤️❤️❤️❤️❤️😓
Mc stan❤❤❤
You know why i am like mc stan.? He selecting old is gold songs always or my Favo songs 😍
DIVINE 🔥🔥🔥
Nice song ❤❤❤
2:37 ❤❤❤❤❤
Kings mc stan 😊😊
Acha h bro aur ho to banaaooo plz bro
Awesome
Thanks
Mc spoil d song ya
Bro I want to purchase the beat, can you provide it .
Awesome sir 🔥❤️
Thanks bro 🥰
@@DIPSROFFICIAL Sir can we do Collab ?
Diven king 👑❤
Given A.P. is 3, 15, 27, 39, …
first term, a = 3
common difference, d = a2 − a1 = 15 − 3 = 12
We know that,
an = a+(n−1)d
Therefore,
a54 = a+(54−1)d
⇒3+(53)(12)
⇒3+636 = 639
a54 = 639+132=771
We have to find the term of this A.P. which is 132 more than a54, i.e.771.
Let nth term be 771.
an = a+(n−1)d
771 = 3+(n −1)12
768 = (n−1)12
(n −1) = 64
n = 65
Therefore, 65th term was 132 more than 54th term.
Or another method is;
Let nth term be 132 more than 54th term.
n = 54 + 132/2
= 54 + 11 = 65th term
12. Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?
Solution:
Let, the first term of two APs be a1 and a2 respectively
And the common difference of these APs be d.
For the first A.P.,we know,
an = a+(n−1)d
Therefore,
a100 = a1+(100−1)d
= a1 + 99d
a1000 = a1+(1000−1)d
a1000 = a1+999d
For second A.P., we know,
an = a+(n−1)d
Therefore,
a100 = a2+(100−1)d
= a2+99d
a1000 = a2+(1000−1)d
= a2+999d
Given that, difference between 100th term of the two APs = 100
Therefore, (a1+99d) − (a2+99d) = 100
a1−a2 = 100……………………………………………………………….. (i)
Difference between 1000th terms of the two APs
(a1+999d) − (a2+999d) = a1−a2
From equation (i),
This difference, a1−a2 = 100
Hence, the difference between 1000th terms of the two A.P. will be 100.
13. How many three digit numbers are divisible by 7?
Solution:
First three-digit number that is divisible by 7 are;
First number = 105
Second number = 105+7 = 112
Third number = 112+7 =119
Therefore, 105, 112, 119, …
All are three digit numbers are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
As we know, the largest possible three-digit number is 999.
When we divide 999 by 7, the remainder will be 5.
Therefore, 999-5 = 994 is the maximum possible three-digit number that is divisible by 7.
Now the series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
first term, a = 105
common difference, d = 7
an = 994
n = ?
As we know,
an = a+(n−1)d
994 = 105+(n−1)7
889 = (n−1)7
(n−1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.
14. How many multiples of 4 lie between 10 and 250?
Solution:
The first multiple of 4 that is greater than 10 is 12.
Next multiple will be 16.
Therefore, the series formed as;
12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows, now;
12, 16, 20, 24, …, 248
Let 248 be the nth term of this A.P.
first term, a = 12
common difference, d = 4
an = 248
As we know,
an = a+(n−1)d
248 = 12+(n-1)×4
236/4 = n-1
59 = n-1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.
15. For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal?
Solution:
Given two APs as; 63, 65, 67,… and 3, 10, 17,….
Taking first AP,
63, 65, 67, …
First term, a = 63
Common difference, d = a2−a1 = 65−63 = 2
We know, nth term of this A.P. = an = a+(n−1)d
an= 63+(n−1)2 = 63+2n−2
an = 61+2n ………………………………………. (i)
Taking second AP,
3, 10, 17, …
First term, a = 3
Common difference, d = a2 − a1 = 10 − 3 = 7
We know that,
nth term of this A.P. = 3+(n−1)7
an = 3+7n−7
an = 7n−4 ……………………………………………………….. (ii)
Given, nth term of these A.P.s are equal to each other.
Equating both these equations, we get,
61+2n = 7n−4
61+4 = 5n
5n = 65
n = 13
Therefore, 13th terms of both these A.P.s are equal to each other.
16. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solutions:
Given,
Third term, a3 = 16
As we know,
a +(3−1)d = 16
a+2d = 16 ………………………………………. (i)
It is given that, 7th term exceeds the 5th term by 12.
a7 − a5 = 12
[a+(7−1)d]−[a +(5−1)d]= 12
(a+6d)−(a+4d) = 12
2d = 12
d = 6
From equation (i), we get,
a+2(6) = 16
a+12 = 16
a = 4
Therefore, A.P. will be4, 10, 16, 22, …
17. Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.
Solution:
Given A.P. is3, 8, 13, …, 253
Common difference, d= 5.
Therefore, we can write the given AP in reverse order as;
253, 248, 243, …, 13, 8, 5
Now for the new AP,
first term, a = 253
and common difference, d = 248 − 253 = −5
n = 20
Therefore, using nth term formula, we get,
a20 = a+(20−1)d
a20 = 253+(19)(−5)
a20 = 253−95
a = 158
Therefore, 20th term from the last term of the AP 3, 8, 13, …, 253.is 158.
18. The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
Solution:
We know that, the nth term of the AP is;
an = a+(n−1)d
a4 = a+(4−1)d
a4 = a+3d
In the same way, we can write,
a8 = a+7d
a6 = a+5d
a10 = a+9d
Given that,
a4+a8 = 24
a+3d+a+7d = 24
2a+10d = 24
a+5d = 12 …………………………………………………… (i)
a6+a10 = 44
a +5d+a+9d = 44
2a+14d = 44
a+7d = 22 …………………………………….. (ii)
On subtracting equation (i) from (ii), we get,
2d = 22 − 12
2d = 10
d = 5
From equation (i), we get,
a+5d = 12
a+5(5) = 12
a+25 = 12
a = −13
a2 = a+d = − 13+5 = −8
a3 = a2+d = − 8+5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3.
19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Solution:
It can be seen from the given question, that the incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP.
Therefore, after 1995, the salaries of each year are;
5000, 5200, 5400, …
Here, first term, a = 5000
and common difference, d = 200
Let after nth year, his salary be Rs 7000.
Therefore, by the nth term formula of AP,
an = a+(n−1) d
7000 = 5000+(n−1)200
200(n−1)= 2000
(n−1) = 10
n = 11
Therefore, in 11th year, his salary will be Rs 7000.
20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.
Solution:
Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.
Hence,
First term, a = 5
and common difference, d = 1.75
Also given,
an = 20.75
Find, n = ?
As we know, by the nth term formula,
an = a+(n−1)d
Therefore,
20.75 = 5+(n -1)×1.75
15.75 = (n -1)×1.75
(n -1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n -1 = 9
n = 10
Hence, n is 10.
Bhai kya tha ye
Bhai 10th math mai question tha
I Love you MCC Stan ❤❤
MC Stan fens😊
Divine always 🔥🔥🔥🔥🔥
Rohit Babu nice to hear ❤❤❤
Chote ache se baat kr😂
😂😂😂😂
My best rapper devine
Big fan ❤❤❤
Super mashup
Din bhar kaam karo raat ko cigarette pite pite rap suno 💀💀
Hii
Mc stan❤❤❤❤❤❤❤
Top hip hop in the wold mc stan🥰🥰🤟🤟