Question

If $25^{th}$ term of an $A.P.$ is $15$ and if its $15^{th}$ term is $25$ , then the $40^{th}$ term of the $A.P.$ is

• A. $-1$
• B. $-10$
• C. $-5$
• D. $0$

Answer 1

Answer: (D)

$0$

Answer 2

Let $a$ be the first term and $d$ be the common difference.
According to question,
$T_{25} = 15$
$\Rightarrow a + 24d = 15\,\,\,...(i)$
$T_{15} = 25$
$\Rightarrow a + 14 \,d = 25\,\,\,...(ii)$
On solving $(i)$ and $(ii)$, we get
$a = 39, d = - 1$
So, $T_{40} = a + 39d = 39 + 39(-1) = 0$