Question

The sum of first ‘n’ terms of an arithemetic progression is given by the formula

Sn

= 3n2 + n, then its 3rd term is​

Answer 1

Step-by-step explanation:

GiVEN

• Sum of first nth term of AP = 3n² + n

TO FiND

• The term 3rd term

SOLUTiON

• 1st term = 3*(1)² + 1
• 1st term = 4

• 2nd term = 3*(2)² + 2 - 4
• 2nd term = 14 - 4
• 2nd term = 10

• 3rd term = 3*(3)² + 3 - 14
• 3rd term = 30 - 14
• 3rd term = 16

Thus , the 3rd term is 16.

Answer 2

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

We know that,

↝ 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.

Now,

According to statement,

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

So,

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

On cancelation of n on both sides,

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

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

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

On comparing with sides,

$\purple{\bf :\longmapsto\:d = 6}$

and

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

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

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

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

$\purple{\bf :\longmapsto\:a = 4}$

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,

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

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

$\rm :\longmapsto\:a_3 = 4 + 2 \times 6$

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

$\red{\bf :\longmapsto\:a_3 = 16}$