Question

Given the AP $10,7,4,.....,-62$. The ${11}^{th}$ term from end of the AP is

• A. $23$
• B. $53$
• C. $34$
• D. $-32$

Answer 1

Answer: (D)

$-32$

Let $d$ be the common difference of the given A.P>

$d=7-10$

$\Rightarrow d=-3$

Here, $a=10,d=-3,t_n=-62$

$n^{th}$ term of an A.P. is given as,

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

Thus,

$-62=10+(n-1)(-3)$

$-72=(n-1)(-3)$

Thus, $n=25$

Hence, in our solution, we are looking for ${11}^{th}$ term from last that is ${15}^{th}$ term from starting.

$t_{15}=10+(15-1)(-3)$

$=-32$

Hence option (D) is correct