Question

If the $7^{th}$ and ${13}^{th}$ terms of an AP are 34 and 64 respectively, then the ${18}^{th}$ term of the AP would be ___.

• A. 87
• B. 88
• C. 89
• D. 90

Answer 1

The correct option is C 89

The formula to find the $n^{th}$ term of an arithmetic progression is $t_n=a+(n-1)d$.

where,

'$a$' is the first term,

'd' is the common difference,

'n' is the no. of terms,

'$t_n$' is the $n^{th}$ term.

Given, $t_7$ = 34
$\Rightarrow$ $a+(7-1)d=34$

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

We have $t_{13}$ = 64

$\Rightarrow$ $a+(13-1)d=64a+12d=64---\left(2\right)$
To find the values of a and d, subtract eqn.(1) from eqn.(2).

$\Rightarrow$ a+12d - (a+6d) = 64 - 34

$\Rightarrow$ 6d = 30

$\Rightarrow$ d = 5

Substituting the value of 'd' in eqn.(1), we get

a+(6 $\times$ 5) =34

$\Rightarrow$ a+ 30 = 34

$\Rightarrow$ a = 4
$\therefore$ ${18}^{th}$ term will be
$t_{18}=4+(18-1)5=89$