# Find the sum of the first 50 common terms of 12,16,20,........ and 18,24,30........ a)19990 b)12700 c)15900 d)18400

#### WIPROkanikannan on 27-Jul-2017 15:43

Find the sum of the first 50 common terms of 12,16,20,........ and 18,24,30........

a)19990
b)12700
c)15900
d)18400
on 28-Jul-2017 12:03

on 28-Jul-2017 12:10

c)15900......... The common terms would be 4*6,4*9,4*12....... Now sum of these numbers would be 4[6+9+12+........upto 50 terms] Sum=4*3*(2+3+4+...........+51) Sum=12*(1325) Sum=15900
on 28-Jul-2017 12:12

this is the problem based on Geometric series first find the common ratio of the gp then use the formula for finding the sum of a GP use S=a(r^n-1)/(r-1) where 'a' is the first number of a gp and 'r' is the common ratio.and 'n' is the no upto which u hav to find the sum
on 28-Jul-2017 17:54

c) 15900
on 29-Jul-2017 12:18

ANS 15900 The common terms would be 4*6,4*9,4*12 sum of these numbers would be 4[6+9+12+................................................................................. to 50 terms] Sum=4*3*(2+3+4+....................................................+51) Sum=12*(1325) Sum=15900
on 29-Jul-2017 12:21

Ans: 15900,
on 29-Jul-2017 14:00

c) 15900
on 29-Jul-2017 15:35

c) 15900
on 29-Jul-2017 22:40

c)15900
on 30-Jul-2017 22:38

By using the formula (n/2)[2a+(n+1)d] and the answer is 15900
on 31-Jul-2017 11:00

The common terms are 24,36,48... This can be written as 4*6,4*9,4*12... Taking 4 common, The sum of these numbers is 4[6+9+12+.... ] till 50 terms Further take 3 common such that Sum=4*3*[2+3+4+...+51] 2,3,4...are in AP with a=2,d=1 Sum of n terms in an AP is 2a+(n-1)d Sum=12*(1325)=15900
on 31-Jul-2017 16:39

17BCE2173
on 31-Jul-2017 18:08

159000
on 31-Jul-2017 23:13

on 1-Aug-2017 14:09

15900
on 1-Aug-2017 14:17

on 1-Aug-2017 22:17

15900
on 3-Aug-2017 08:42

ha
on 3-Aug-2017 16:25

option c 15900
on 4-Aug-2017 00:13

15900
on 4-Aug-2017 08:38

c .15900 used concept of arthematic progression formula used n/2[2a+(n-1)d]
on 4-Aug-2017 09:21

c.15900
on 4-Aug-2017 13:42

The common terms are 24,36,48... This can be written as 4*6,4*9,4*12... Taking 4 common, The sum of these numbers is 4[6+9+12+.... ] till 50 terms Further take 3 common such that Sum=4*3*[2+3+4+...+51] 2,3,4...are in AP with a=2,d=1 Sum of n terms in an AP is 2a+(n-1)d Sum=12*(1325)=15900
on 4-Aug-2017 17:49

159000
on 5-Aug-2017 11:56

c) 159000
on 5-Aug-2017 12:30

The common terms will be multiples of both 6 and 4. LCM(6,4)=12 sum of 50 terms= 25(24*2+(49*12))=159000
on 5-Aug-2017 15:43

1590000
on 5-Aug-2017 23:37

on 6-Aug-2017 13:46

24,36,48->common terms it can be written as 4*6,4*9,4*12 taking 4 common->4*3[2+3+4..50] 2,3,4 can be found using AP->>a=2,d=1 formula:2a+(n-1)d=12*(1325)=15900
on 7-Aug-2017 11:52

15900
on 7-Aug-2017 15:38

It is AP series. Therefore, by applying sn formula we get 15900
on 8-Aug-2017 21:00

the both series is in AP series we can find the nth term of series and we can get solution c)15900
on 9-Aug-2017 14:56

on 10-Aug-2017 14:09

on 11-Aug-2017 14:14

The common terms are 24,36,48... This can be written as 4*6,4*9,4*12... Taking 4 common, The sum of these numbers is 4[6+9+12+.... ] till 50 terms Further take 3 common such that Sum=4*3*[2+3+4+...+51] 2,3,4...are in AP with a=2,d=1 Sum of n terms in an AP is 2a+(n-1)d Sum=12*(1325)=15900
on 11-Aug-2017 20:18

A.P with a = 24 and d = 12. sum of the first n-terms of A.P = (n/2)[2a + (n-1)d] n = 50 and required sum = (50/2)*[2(24) + 49(12)] = 25*[636]= 15900. the answer is 15900.
on 11-Aug-2017 20:54

Series1---->12,16,20,24,... and common difference = 4 Series2 ---->18,24,30,36...and common difference =6 the LCM between 4 and 6 =12 so it is 24,36,48,..... to find 50th term=a+(n-1)*d and a--->first term---->24 d--->difference-->12 and n-->50 50th term=24+(50-1)*12 =24+(49)*12 =24+588 50th term=612 Sum of 50 common terms=n/2(a+L) ,L--->last term(50th term)--->612,a--->24,n-->50 Sum of 50 common terms=50/2(24+612) =25(636)=15900 ANSWER ----->15900
on 14-Aug-2017 16:29

Series1---->12,16,20,24,... and common difference = 4 Series2 ---->18,24,30,36...and common difference =6 the LCM between 4 and 6 =12 so it is 24,36,48,..... to find 50th term=a+(n-1)*d and a--->first term---->24 d--->difference-->12 and n-->50 50th term=24+(50-1)*12 =24+(49)*12 =24+588 50th term=612 Sum of 50 common terms=n/2(a+L) ,L--->last term(50th term)--->612,a--->24,n-->50 Sum of 50 common terms=50/2(24+612) =25(636)=15900 ANSWER ----->15900
on 14-Aug-2017 16:33

on 15-Aug-2017 18:15

15900
on 15-Aug-2017 21:58

C) 15900 The common terms would be 4*6,4*9,4*12... Taking 4 as common, 4(6+9+12...) upto 50 terms since they are multiples of 3's, taking 3 as common 4*3*(2+3+...+51) it is in AP, a=2,d=1. (n/2)[2a+(n-1)d ]= 1325 ANS: 1325*12 = 15900
on 16-Aug-2017 17:36

12700
on 16-Aug-2017 18:58

First sequence is multiples of 4 starting from 12 while the second one is multiples of 6 starting from 18 L.C.M. of 4 and 6 is 12. So the common terms will be the multiples of 12 starting from 24 And the sequence will be as 24,36,48,............50 terms sum of terms=24+36+48+.......50terms sum=12(2+3+4+......+51) consider 2+3+4+.....+51 Sn=(n/2)(2a+(n-1)d) where n=>no.of terms, a=>first term, d=>common difference here n=50,a=2,d=1 On simplification Sn=1325 Therefore required sum =12*1325 =>Required sum=15900
on 17-Aug-2017 09:58

A.P with a = 24 and d = 12. sum of the first n-terms of A.P = (n/2)[2a + (n-1)d] n = 50 and required sum = (50/2)*[2(24) + 49(12)] = 25*[636]= 15900. the answer is 15900.
on 17-Aug-2017 19:00

15900