Question

if the sum of first 10 terms of an a.p is 140 and the sum of first 16 terms is 320. find the sum of n terms.

Answer 1

hence Sn=4n² as the sum of n term

Answer 2

Answer:

The Sum of n terms is n² + 4n

Step-by-step explanation:

Given: Sum of first 10 terms of an AP = 140

          Sum of first 16 term of an AP = 320

To find: Sun of n terms

Sum of n terms of an AP is given by,

$S_n=\frac{n}{2}(2a+(n-1)d)$

here a is first term and d is common difference

from given info,

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

$140=5(2a+9d)$

$2a+9d=\frac{140}{5}$

$2a+9d=28$ .............. (1)

$S_{16}=\frac{16}{2}(2a+(16-1)d)$

$320=8(2a+15d)$

$2a+15d=\frac{320}{8}$

$2a+15d=40$ .............. (2)

Subtract eqn (1) from (2) we get,

(2a + 15d) - (2a + 9d) = 40 - 28

2a + 15d - 2a -9d = 12

6d = 12

d = 2

Now put this value in eqn (1), we get

2a + 9×2 = 28

2a + 18 = 28

2a = 28 - 18

2a = 10

a = 5

So, Sum of n term , $S_n=\frac{n}{2}(2\times5+(n-1)\times2)$

                                 $S_n=\frac{n}{2}(10+2n-2)$

                                 $S_n=n(5+n-1)$

                                 $S_n=n(4+n)$

                                 $S_n=n^2+4n$

Therefore, The Sum of n terms is n² + 4n