Question

If the sum of 'n' terms of an A.P is 3n+5n and a = 164,
then n =​

Answer 1

Appropriate Question :-

$\sf \: If \: the \: sum \: of \: n \: terms \: of \: an \: AP \: is \: {3n}^{2} + 5n \: and \:$

$\sf \: a_n \: = 164 \: , then \: n \: = \:$

Solution :-

Given that,

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

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 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.

Tʜᴜs,

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

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

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

☆ On comparing both sides, we get

$\rm :\longmapsto\:d = 6 \:$

and

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

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

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

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

$\bf\implies \:a = 8$

Now,

It is given that

$\rm :\longmapsto\:a_n \: = \: 164$

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,

$\rm :\longmapsto\:164 = 8+ (n - 1)6$

$\rm :\longmapsto\:164 - 8 = (n - 1)6$

$\rm :\longmapsto\:156= (n - 1)6$

$\rm :\longmapsto\:26= (n - 1)$

$\bf\implies \:n \: = \: 27$