Question

Find the sum of all the 11 terms of an AP whose middle terms is 30.


No spam!! ​

Answer 1

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

Given that

• An AP series consists of 11 terms whose middle term is 30.

Since, Number of terms, n = 11

So, there is only 1 middle term.

$\rm\implies \:Middle \: term = {\bigg[\dfrac{11 + 1}{2} \bigg]}^{th} = {6}^{th} \: term$

So, it means

$\rm :\longmapsto\:a_6 = 30$

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,

$\rm :\longmapsto\:a + (6 - 1)d = 30$

$\rm\implies \:\boxed{\tt{ a + 5d = 30}} - - - - (1)$

Now,

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.

Thus,

$\rm :\longmapsto\:S_{11} = \dfrac{11}{2}\bigg(2a + (11 - 1)d \bigg)$

$\rm :\longmapsto\:S_{11} = \dfrac{11}{2}\bigg(2a + 10d \bigg)$

$\rm :\longmapsto\:S_{11} = \dfrac{11}{2} \times 2 \times \bigg(a + 5d \bigg)$

$\rm :\longmapsto\:S_{11} = 11 \times 30$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (1) \bigg\}}$

$\rm\implies \:\boxed{\tt{ \: \: \: \: \: \large{ S_{11} \: = \: 330 }\: \: \: \: \: \: }}$