Question

Let the sum of the first three terms of an $A.P.$ be $39$ and the sum of its last four terms be $178$. If the first term of this $A.P.$ is $10$, then the median of the $A.P.$ is :

• A. 26.5
• B. 28
• C. 29.5
• D. 31

Answer 1

Answer: (C)

Let $a-d, a, a+d$ be three terms in A.P.
$\Rightarrow (a-d)+a+(a +d)=39$
$\Rightarrow a=13 \Rightarrow (a-d)=$ Ist term $=10$ (given)
$\Rightarrow 13-d=10 \Rightarrow d=3$
$\Rightarrow $ Given AP is $10,13,16, \ldots .$
Let there be $n$ terms in AP
$\Rightarrow $ Sum of last four terms $=178$ (given)
$\Rightarrow \frac{4}{2}(2 l+(4-1)(-3))=178$
$\Rightarrow 2(21-9)=178$
$\Rightarrow 4 l =196$
$\Rightarrow 1=49=t_{n}$
$\Rightarrow 10+(n-1)(3)=49$
$\Rightarrow 3 n=42$
$\Rightarrow n=14$
$\therefore $ Median of $AP =\frac{\frac{n}{2} \text { th term }+\left(\frac{n}{2}+1\right) \text { th term }}{2}$
$=\frac{7 \text { th term }+\text { 8th term }}{2}$
$=\frac{t_{7}+t_{8}}{2}=\frac{t_{1}+t_{14}}{2}=\frac{10+49}{2}=\frac{59}{2}=29.5$