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. $29.5$
• B. $28$
• C. $31$
• D. $26.5$

Answer 1

Answer: (A)

$29.5$

Let the first three terms of the A.P. be $a-d,a,a+d$
Given $a-d=10...\left(1\right)$
Also given $a-d+a+a+d=39$
$\Rightarrow 3a=39...\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get
$a=13,d=3$
So, the A.P. is $10,13,16,...$
Now, let the last four terms be $a_n,a_{n-1},a_{n-2},a_{n-3}$
$a+(n-1)d+a+(n-2)d+a+(n-3)d+a+(n-4)d=178$
$\Rightarrow 4a+(4n-10)d=178$
$\Rightarrow n=14$
So, total number of terms in the A.P. are $14$ (even).
Hence, median of the A.P. will be mean of $T_7,T_8$
$T_7=10+6\left(3\right)=28$
$T_8=10+7\left(3\right)=31$
Median $=\frac{28+31}{2}=29.5$