Question

.Sn
of AP is 3n2 + 4n then 8th term is

Answer 1

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

Given that

Sum of n terms of an AP series is given by

$\rm :\longmapsto\:S_n \: = \: {3n}^{2} + 4n$

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,

$\rm :\longmapsto\:\dfrac{n}{2} \bigg(2a + (n - 1)d \bigg) = n(3n + 4)$

$\rm :\longmapsto\:2a + (n - 1)d = 6n + 8$

$\rm :\longmapsto\:2a + nd - d = 6n + 8$

$\rm :\longmapsto\:nd + (2a- d) = 6n + 8$

So, On comparing, we get

$\bf\implies \:d = 6$

Also,

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

$\rm :\longmapsto\:2a - 6= 8$

$\rm :\longmapsto\:2a= 8 + 6$

$\rm :\longmapsto\:2a= 14$

$\bf\implies \:a = 7$

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,

$\purple{\rm :\longmapsto\:a_8}$

$\rm \: = \: a + 7d$

$\rm \: = \: 7 + 7 \times 6$

$\rm \: = \: 7 + 42$

$\rm \: = \:49$

Thus,

$\purple{\rm :\longmapsto\:\boxed{ \: \: \: \: \: \large{\tt{ a_8 = 49}} \: \: \: \: \: \: }}$