Question

if sum of first n terms of an Ap is Sn=3n^2+n, then its 3rd term is?​

Answer 1

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

Given that

↝ Sum of the first n terms of an AP is

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

So, on substituting the value, we get

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

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

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

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

↝ On comparing both sides, we get

$\bf\implies \:d = 6$

and

$\rm :\longmapsto\:2a - d = 2$

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

$\rm :\longmapsto\:2a = 2 + 6$

$\rm :\longmapsto\:2a = 8$

$\bf\implies \:a \: = \: 4$

Now,

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

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.

Tʜᴜs,

↝ 3ᵗʰ term is,

$\rm :\longmapsto\:a_3 = a + (3 - 1)d$

$\rm :\longmapsto\:a_3 = a + 2d$

↝ On substituting the values of a and d, we get

$\rm :\longmapsto\:a_3 = 4 + 2(6)$

$\rm :\longmapsto\:a_3 = 4 +12$

$\bf :\longmapsto\:a_3 = 16$