Question

If the ${14}^{th}$ term of an arithmetic series is $6$ and $6^{th}$ term is $14$, then what is the ${95}^{th}$ term?

• A. $-75$
• B. $75$
• C. $80$
• D. $-80$

Answer 1

The correct option is A $-75$

Given, ${14}^{th}$ term $=6$, $6^{th}$ term $=14$

We know $T_n=a+(n-1)d$

Therefore, $6=a+(14-1)d=a+13d$ ....(1)
and $14=a+(6-1)d=a+5d$ ....(2)

Subtracting equations (1) and (2),

$d=8d=-8\Rightarrow d=-1$
Putting this value in equation (1), we get

$6=a+13(-1)$

$\Rightarrow a=6+13$

$\Rightarrow a=19$
Thus $T_{95}=a+(95-1)d$
$=a+94d$
$=19-94=-75$