Question

an ap 12th term is - 13 and the sum of of first 4 terms is 24find the sum of drst 10 terms

Answer 1

Given Question:-

• An AP whose 12th term is - 13 and the sum of the first 4 terms is 24. Find the sum of first 10 terms.

─━─━─━─━─━─━─━─━─━─━─━─━─

Answer :-

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{a_{12} = - 13} \\ &\sf{S_4 \: = 24} \end{cases}\end{gathered}\end{gathered}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{S_{10}} \end{cases}\end{gathered}\end{gathered}$

─━─━─━─━─━─━─━─━─━─━─━─━─

Formula used:-

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\blue{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \: ( \: 2\:a\:+\:(n\:-\:1)\:d \: )}}}}}} \\ \end{gathered}$

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.

• Sₙ is sum of first n terms.

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\purple{\bold{Solution :- }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\begin{gathered}\bf Let - \begin{cases} &\sf{first \: term \: of \: an \: ap \: be \: a} \\ &\sf{common \: difference \: be \: d} \end{cases}\end{gathered}\end{gathered}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\red{\bold{❥︎Step :- 1 }}$

☆ Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\:\:{{{{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

➣ It is given that 12ᵗʰ term is - 13

$\bf\implies \: - 1 3 \: = \: a \: + \: 11 \: d$

$\bf\implies \:a \: = - 13 \: - \: 11d\: - - - (i)$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\red{\bold{❥︎Step :- 2 }}$

➣ It is given that sum of first 4 terms is 24.

Using formula,

$\bf \:  ⟼ S_n\:=\dfrac{n}{2} \: ( \: 2\:a\:+\:(n\:-\:1)\:d \: )$

$\sf \:  ⟼ \: 24 \: = \: \dfrac{4}{2} (2a + (4 - 1) \times d)$

$\sf \:  ⟼12 \: = 2a + 3d$

$\sf \:  ⟼12 = 2( - 13 - 11d) + 3d \: \: \: \: \: \: \: \: \: (using \: (i))$

$\sf \:  ⟼ \: 12 \: = \: - \: 26 \: - \: 22 \: d \: + \: 3 \: d$

$\sf \:  ⟼ \: - 19 \: d \: = \: 38$

$\bf\implies \: \boxed{d \: = \: - \: 2}$

☆ On substituting d = - 2 in equation (i), we get

$\sf \:  ⟼ \: a \: = \: - \: 13 \: - 11 \: \times \: ( - \: 2)$

$\sf \:  ⟼ \: a \: = - 13 + 22$

$\bf\implies \: \boxed{a \: = \: 9}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\red{\bold{❥︎Step :- 3 }}$

☆ To find sum of 10 terms

$\bf \:  ⟼ S_{10}\:=\dfrac{10}{2} \: ( \: 2\: \times 9\:+\:(10\:-\:1)\:( - 2) \: )$

$\bf \:  ⟼ S_{10}\:=\dfrac{10}{2} \: ( \: 18 \: - \: 18)$

$\bf\implies \: \boxed{S_{10} \: = \: 0}$

─━─━─━─━─━─━─━─━─━─━─━─━─