Question

In an A.P, Sn = 2n2 + 6n, Find the First term and common difference.

Answer 1

Answer:

Sn=2n^2+6n

S1=2.1^2+6.1=2.1+6=2+6=8

S1 is the 1st term=8

S2=2.2^2+6.2=2.4+12=8+12=20

S2 is the sum of 1st and 2nd term

1st term=8

so 2nd term =20-8=12

so common difference=12-8=4

so 1st term is 8 and common difference is 4

Answer 2

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

Given that, Sum of first n terms of an AP is

$\red{\rm :\longmapsto\:S_n = {2n}^{2} + 6n}$

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,

$\rm :\longmapsto\: {2n}^{2} + 6n \: = \: \dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)$

$\rm :\longmapsto\: n({2n} + 6)\: = \: \dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)$

$\rm :\longmapsto\: {2n} + 6\: = \: \dfrac{1}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)$

$\rm :\longmapsto\: {4n} + 12\: = \: 2 \:a\:+\:(n\:-\:1)\:d$

$\rm :\longmapsto\: {4n} + 12\: = \: 2 \:a\:+\:nd\:-\:d$

$\rm :\longmapsto\: {4n} + 12\: = \: (2 \:a\: - \: d \: ) \: +\:nd$

So, on comparing we get

$\rm :\longmapsto\:d = 4$

and

$\rm :\longmapsto\:2a - d = 12$

$\rm :\longmapsto\:2a - 4 = 12$

$\rm :\longmapsto\:2a = 12 + 4$

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

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

Hence,

First term of an AP = 8

and

Common Difference of an AP = 4

Additional Information :-

↝ 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.