Question

36. The sum of 1st 12 terms of an AP is 144 and the sum of next two terms
is 52. Find
a. 20th term
b. Sum upto 20 terms​

Answer 1

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

Given that,

• Sum of first 12 terms of an AP = 144

We know,

↝ Sum of first 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 first n terms.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Tʜᴜs,

$\rm :\longmapsto\:S_{12}\:=\dfrac{12}{2} \bigg(2\:a\:+\:(12\:-\:1)\:d \bigg)$

$\rm :\longmapsto\:144\:= 6 \bigg(2\:a\:+\:(11)\:d \bigg)$

$\bf :\longmapsto\:24\:= 2\:a\:+\:11\:d - - - (1)$

According to statement,

• Sum of 13ᵗʰ and 14ᵗʰ term is 52

We know that

↝ 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,

$\rm :\longmapsto\:a_{13} + a_{14} = 52$

$\rm :\longmapsto\:a + 12d + a + 13d = 52$

$\bf :\longmapsto\:2a + 25d = 52 - - - (2)$

On Subtracting equation (1) from equation (2), we get

$\rm :\longmapsto\:2a + 25d - 2a - 11d = 52 - 24$

$\rm :\longmapsto\:14d = 28$

$\bf\implies \:d = 2$

On substituting d = 2 in equation (1), we get

$\rm :\longmapsto\:2a + 11 \times 2 = 24$

$\rm :\longmapsto\:2a + 22 = 24$

$\rm :\longmapsto\:2a = 24 - 22$

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

$\bf\implies \:a = 1$

Hence,

$1. \: \: \: \red{\boxed{ \bf \: {20}^{th} \: term \: of \: an \: AP}}$

$\rm :\longmapsto\:a_{20} = a + (20 - 1)d$

$\rm :\longmapsto\:a_{20} = a + 19d$

$\rm :\longmapsto\:a_{20} = 1 + 19 \times 2$

$\rm :\longmapsto\:a_{20} = 1 +38$

$\bf :\longmapsto\:a_{20} = 39$

$2. \: \: \: \red{\boxed{ \bf \: Sum \: of \: first \: {20} \: terms \: of \: an \: AP}}$

$\rm :\longmapsto\:S_{20}\:=\dfrac{20}{2} \bigg(2\:a\:+\:(20\:-\:1)\:d \bigg)$

$\rm :\longmapsto\:S_{20}\:=10 \bigg(2\:a\:+\:19\:d \bigg)$

$\rm :\longmapsto\:S_{20}\:=10 \bigg(2\: \times 1\:+\:19\: \times 2 \bigg)$

$\rm :\longmapsto\:S_{20}\:=10 \bigg(2 + 38 \bigg)$

$\rm :\longmapsto\:S_{20}\:=10 \bigg(40 \bigg)$

$\bf :\longmapsto\:S_{20}\:=400$