Question

Find the A.P whose $7^{th}$ and ${13}^{th}$ terms are respectively $34$ and $64$.

Answer 1

As we know that $n^{th}$ term of an A.P. is-

$a_n=a+(n-1)d$

where as,

$a=$ First term of A.P.

$d=$ comon difference

Therefore,

$a_7=34\left[\text{Given}\right]$

$a+6d=34$

$\Rightarrow a=34-6d.....\left(1\right)$

$a_{13}=64\left[\text{Given}\right]$

$a+12d=64$

$a=64-12d.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$34-6d=64-12d$

$6d=30\Rightarrow d=5$

Substituting the value of $d$ in ${eq}^n\left(1\right)$, we have

$a=34-6\times 5=4$

$\therefore$ The first term and common difference of A.P. are $4$ and $5$ respectively.

Hence the A.P. is $4,9,14,........$.