Question

the sum of n terms of an ap. is 3nsquarw +5n Find the ap hence find its 16th term
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Answer 1

Appropriate Question :-

If the sum of n terms of an AP series is 3n² + 5n. Find the AP and hence find its 16th term.

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

Given that, Sum of n terms of an AP is

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

Let assume that

First term of an AP = a

Common difference of an AP = d

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 this value, we get

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

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

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

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

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

So, on comparing we get

$\rm \: d = 6$

and

$\rm \: 2a - d = 10$

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

$\rm \: 2a = 10 + 6$

$\rm \: 2a = 16$

$\rm\implies \:a = 8$

So, required AP series is 8, 14, 20, 26, ...

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,

$\rm \: a_{16}$

$\rm \: = \: a + (16 - 1)d$

$\rm \: = \: a + 15d$

$\rm \: = \: 8 + 15 \times 6$

$\rm \: = \: 8 + 90$

$\rm \: = \: 98$

Hence,

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