Question

Determine the $n^{th}$ term of the AP whose 7th term is −1 and 16th term is 17.

Answer 1

Let a be the first term and d be the common difference of the AP. Then,
$a_7=-1$

$\Rightarrow a+(7-1)d=-1$ [$\because n^{th}$ term is given by $a_n=a+(n-1)d$]

$\Rightarrow a+6d=-1\dots \dots \left(i\right)$

Also, $a_{16}=17$

$\Rightarrow a+15d=17\dots \dots \left(ii\right)$

From eq. (i) and (ii), we get
$-1-6d+15d=17$ [$\because a=-1-6d$ from eq.(i)]

$\Rightarrow 9d=17+1$

$\Rightarrow d=2$
Putting $d=2$ in $eq.\left(i\right),$ we get
$a+6\times 2=-1$

$\Rightarrow a=-1-12=-13$
$\therefore a_n=a+(n-1)d=-13+(n-1)\times 2$

$=-13+2n-2=2n-15$
Hence, the $n^{th}$ term of the AP is $(2n-15).$