Fallin for you (slowed+reverb) Shrey Singhal 

Voice was just.................❤

Nice song 😊

This is such a amazing song l feel better when I listen this song....... ❤.........❤❤........

This song 💖 amizing and loyal boyz feel best __most romantic 🤩 boyz match this song 🎉😂(fabulous 😍 song 💖 nyc nyc editing ❤️

Uff my fav❤

2:11 this line was my favourite and this song is heaven if this song sing for me ❤❤

awesome song

This song always bring those loveable memory with him 😊 , he is still with me and hope always be ❤😊 , this song is really much amazing 🎉... Admit or not but i'm sure there must be one person who is coming in ur mind whlile listening it right ???😊😊

Fav song ❤❤.

I really fallin this song waves

1:45 huhh favorite part❤❤

Nice💗💗💗

This song best yarr my Fevaret ❤❤❤❤ nice 👌😍

I love this song ❤️❤️❤️❤️❤️❤️❤️

Fallin for you Jo he hamara vo hamse break up krna chata h or hum uske piche lage hue he 😢😢😢😢😢😢 ❤❤❤❤ i miss you my chand 🌜🌜

2:12 .....2:35 best part of this song😍🤩

❤this song my favorite song

really "fallin for this song

imagine when your partner with you and suddenly this song activate ❤😮

0:14

This song sing by a boy for a girl..May this song will sing by someone for you girls ...❣️❣️

Such a attitudnal song.....😉🤗

❤❤❤❤❤❤ nice song 😻😻

Listen this song fell attitude and confidential

I love singer singing

Literally I Fallin in someone's love

2:12 to 2:35 is best part

❤❤

Love😂😂😔 real love available only 4% 😌🥀💯💯

💜💜

👀💖🙃

Feel the song for your first love at first meeting ❤

bohat accha tah aa song

Littrally I Love This Song ❤

❤️❤️❤️❤️❤️❤️love this song🥰🥰🥰🥰🥰🥰🥰🥰🥰😭😭

❤❤❤❤❤❤❤❤❤

Nice 🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉

beautiful song

Super nice video

song is good,s😉

🖤🖤

I love.this.song❤❤❤❤❤❤😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊

I am falling your love ❤❤

2024❤❤

😘😘😘🤗🤗🤗🤗😘😘

I called my bae Baby i am falling for you she smiled ❤

nixe🥰🥰🥰🥰🥰🙃🙃

NICE

nice song bast 😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍

😮😮😮😮

Starting💀💀

❤❤❤❤❤❤❤

onesong is enough for monetization of youtube channel

Love this song 💖

ye gana sun kr hayat or murat ki love story yaad aati ha

Anyone 2024❤

😘😘😘

😊😊😊😊

Lvl ❤

l have no time for commet

hyeeeeee

But I enjoy this song

ap ya gana tab suna jab ap ka result ayga or phir papa say teray kharchay ya sanda hai uper say to ya result la kar deta hai 😂🤣😂🤣🤣🤣🤣🤣

This for you 😊

11. Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54th term? Solution: 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.

hey

broke😀😀

🎉

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Hey god pls send a right person for inocent and introvert girl.. 😂👾

গানটাঅনেকভোলাগেআমারমনছুযযায়রেগো❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤

BTS I love you so 💝💝💝😔😔💝💝😔😔😊😊😔😔😊😊😔😔

🥰🥰🥰🥰🥰

I only love shajiza hospital. .others only use for making trap..tum sabka istemal kiya gya hai . .

Lora vich 😂😂😂😂

BTS💋💋💋💋💋💋💋🙌🙌💘😍😍😍😍💏💑VVJk

Nice song 😍

Nice❤❤❤❤

❤❤

BTS I love you so 💝💝💝😔😔💝💝😔😔😊😊😔😔😊😊😔😔

BTS I love you so 💝💝💝😔😔💝💝😔😔😊😊😔😔😊😊😔😔