Mathematics - Sequence & Series Question with Solution | TestHub

Let the A.P. be$a, a+d,a+2d\cdots \cdots ,l-3d, l-2d, l-d,l$$$
Where$a$is the first term and$l$is the last term,
Sum of$1^{st}$,$3$terms is,$39$.
$30+3d=39$as$a=10$(Given)
$d=\frac{9}{3}=3$
Sum of last$4$terms is$178$.
$4l-6d=178$
$l=49$
$10, 13, 16, 19.......46, 49$
Let number of terms are,$n$then$n^{th}$term is$10 + (n - 1) 3 = 49$
$\Rightarrow n = 14$
So the median of the series would be mean of$7^{th} and 8^{th}$term
$\frac{10+6\cdot \left(3\right)+10+7\cdot 3}{2}$
$=\frac{59}{2} =29.5$

Alternate way
The median would be mean of 10 and 49, That is 29.5.