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 $10,13,16,\dots .$
Let there be $n$ terms in AP
$\Rightarrow$ Sum of last four terms $=178$ (given)
$\Rightarrow \frac{4}{2}(2l+(4-1)(-3))=178$
$\Rightarrow 2(21-9)=178$
$\Rightarrow 4l=196$
$\Rightarrow 1=49=t_n$
$\Rightarrow 10+(n-1)(3)=49$
$\Rightarrow 3n=42$
$\Rightarrow n=14$
$\therefore$ Median of $AP=\frac{\frac{n}{2}\text{ th term }+\left(\frac{n}{2}+1\right)\text{ th term }}{2}$
$=\frac{7\text{ th term }+\text{ 8th term }}{2}$
$=\frac{t_7+t_8}{2}=\frac{t_1+t_{14}}{2}=\frac{10+49}{2}=\frac{59}{2}=29.5$