Question

15. If the sum of first n terms of a series in A.P. is n^2+3n, find the 10thterm.
b) 16
a) 10
c) 18
d) 22​

Answer 1

Answer:-

The Answer Is Option d) 22.

Explanation:-

Given:-

• Sum of first n terms of an AP is n² + 3n.

To Find:-

• The 10th term of the AP.

Formula Used:-

$\\ \large\orange{ \mapsto \boxed{ \purple{ \bf{a_n = (S_n) - (S_{n - 1}).}}}}$

Where,

• $\tt a_n = n^{th}$ Term.

• $\tt S_n$ = Sum of n terms.

Therefore,

$\\ \orange{ \mapsto \boxed{ \pink{ \bf{a_{10} = (S_{10}) - (S_{9}).}}}}$

Also,

$\\ \tt \mapsto S_n = {n}^{2} + 3n. \: \: \: \: \bigg( \bf given \bigg).$

And Hence,

$\\ \tt \mapsto S_{10} = {10}^{2} + 3(10).$

$\\ \tt = 100 + 30.$

$\\ \large\bf = 130.$

Similarly,

$\\ \tt \mapsto S_{9} = {9}^{2} + 3(9).$

$\\ \tt = 81 + 27.$

$\\ \large \bf = 108.$

Therefore,

$\\ \bf\mapsto a_{10} = (S_{10}) - (S_{9}).$

$\\ \tt\mapsto a_{10} =130 - 108.$

$\large \orange{\mapsto \boxed{ \blue{ \bf a_{10} =22.}}}$

Therefore 10th term is 22.

So The Final Answer is Option (d).

Answer 2

Given ,

The sum of first n terms of a series in AP is (n)² + 3n

As we know that ,

$\boxed{ \tt{s_{1} = a_{1} }}$

And

$\boxed{ \tt{a_{n} = s_{n} - s_{n - 1} }}$

Thus ,

First term = (1)² + 3(1) = 4

Second term = 4 - (2)² + 3(2) = 6

∴Common difference (d) = 2

Now , the nth term of an AP is given by

$\boxed{ \tt{ a_{n} = a + (n - 1)d}}$

Thus ,

10th term = 4 + (10 - 1)2

10th term = 4 + 18

10th term = 22

∴ The 10th term of given AP is 22