The sum of the first five terms of an AP is 55 and the sum of the first ten terms of this AP is 235, find the sum of its first 20 terms
Answers
Answered by
65
use formula
Sn=n/2 {2a+(n-1) d}
now
55=5/2 {2a+(5-1) d}
110=10a+20d
a+2d=11 ------------(1)
again question ask
235=10/2 {2a+(10-1) d}
235=10a+45d
2a+9d=47----------(2)
solve (1)and (2) equation
a=1 and d =5
now
S20=20/2 {2 x 1+(20-1) x 5}
= 10 (2+95)=970
Sn=n/2 {2a+(n-1) d}
now
55=5/2 {2a+(5-1) d}
110=10a+20d
a+2d=11 ------------(1)
again question ask
235=10/2 {2a+(10-1) d}
235=10a+45d
2a+9d=47----------(2)
solve (1)and (2) equation
a=1 and d =5
now
S20=20/2 {2 x 1+(20-1) x 5}
= 10 (2+95)=970
Answered by
5
Let S
n
be the sum of n terms of an AP with first term a and common difference d. Then
S
n
=
2
n
(2a+(n−1)d)
putting n=5 in S
n
, we get
S
5
=
2
5
(2a+4d)⇒S
5
=5(a+2d) ...(1)
putting n=7 in S
n
, we get
S
7
=
2
7
(2a+6d)⇒S
7
=7(a+3d) ...(2)
Adding (1) and (2), we get
12a+31d=167 ...(3)
Now,S
10
=235⇒
2
10
(2a+9d)=235
⇒2a+9d=47 ...(4)
On applying 6×(4)−(3), we get
23d=115⇒d=5
Putting d=5 in (3), we get
12a=167−31×5⇒a=1
Therefore, Required sum S
20
=
2
20
[2×1+(20−1)×5
⇒S
20
=10[2+95]
⇒S
20
=970
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