Question

In an ap, the sum of n terms of ap is 3n^2 + 5n. Find the AP, hence find its 16th term.​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:An \: AP \: series \: having \: S_n = {3n}^{2} + 5n$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:An \: AP \: series \: and \: a_{16} \: term$

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

↝ nᵗʰ term of an arithmetic sequence is,

$\rm :\longmapsto\:\begin{gathered}\:\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

And

↝ Sum of 'n' terms of an arithmetic sequence is,

$\rm :\longmapsto\: S_{n} = \dfrac{n}{2} \bigg(2a + (n - 1)d \bigg)$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

• Sₙ is the sum of first 'n' terms.

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:S_n = {3n}^{2} + 5n$

$\rm :\longmapsto\: \dfrac{\cancel{n}}{2} \bigg(2a + (n - 1)d \bigg) = \cancel{n}(3n + 5)$

$\rm :\longmapsto\:2a + (n - 1)d = 6n + 10$

$\rm :\longmapsto\:2a + nd - d = 6n + 10$

$\rm :\longmapsto\:(2a - d) + nd = 6n + 10$

On comparing, we get

$\rm :\longmapsto\:d = 6 \: \: and \: 2a - d = 10$

$\rm :\longmapsto\:2a - 6 = 10$

$\rm :\longmapsto\:2a = 16$

$\rm :\implies\:a = 8$

Hence,

$\rm :\longmapsto\:An \: AP \: series \: is \: 8, \: 14, \: 20, \: 26, \: - - -$

Now,

$\rm :\longmapsto\:a_{16} = a + (16 - 1)d$

$\rm :\longmapsto\:a_{16} =a + 15d$

$\rm :\longmapsto\:a_{16} =8 + 15 \times 6$

$\rm :\longmapsto\:a_{16} =8 + 90$

$\rm :\longmapsto\:a_{16} =98$