Question

the sum of first 20 terms of an ap is 400 and sum of first 40 term is 1600 find the sum of its 10 terms

Answer 1

let the first 20terms
s20=20/2×(2a+19d)
400=20/2(2a+19d)
400=10[2a+19d]
2a+19d=40.......1
now again
s40=40/2(2a+39d)
1600=20[2a+39d]
2a+39d=80....2
from 1 and 2 , we get
a=1 and d=2
s10 =10/2 [2×1+(10-1)(2)]
=5[2+9×2]
=5[2+18]
=5×20
=100


it will help you...

Answer 2

Answer:

100

Step-by-step explanation:

Formula of sum of first n terms in A.P.=$S_n =\frac{n}{2}(2a+(n-1)d)$ ----1

Put n = 20

$S_{20} =\frac{20}{2}(2a+(20-1)d)$

We are given that the sum of first 20 terms of an A.P. is 400

$400 =10(2a+19d)$

$400 =20a+190d$

$40 =2a+19d$  --2

Put n = 40 in 1

$S_{40} =\frac{40}{2}(2a+(40-1)d)$

$S_{40} =20(2a+39d)$

We are given that sum of first 40 term is 1600

$1600 =40a+780d$

$160 =4a+78d$

$80 =2a+39d$  --3

Solve 2 and 3

Subtract a from b

$80 -40 =2a+39d-2a-19d$

$40 =39d-19d$

$40 =20d$

$\frac{40}{20} =d$

$2 =d$

Substitute the value of d in 2

$40 =2a+19(2)$

$40 =2a+38$

$40-38=2a$

$2=2a$

$1=a$

So, a = 1

d =2

So, to find the sum of first 10 terms

Substitute n = 10 in 1

$S_{10} =\frac{10}{2}(2(1)+(10-1)2)$

$S_{10} =5(2+18)$

$S_{10} =100$

Hence the sum of first 10 terms is 100