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. $26.5$
• C. $28$
• D. $41$

Answer 1

The correct option is A $29.5$

$10+(10+d)+(10+2d)=39\Rightarrow d=3\Rightarrow t_n=10+(n-1)d\Rightarrow t_n=3n+7$

Now,
$t_n+t_{n-1}+t_{n-2}+t_{n-3}=178\Rightarrow 7\times 4+3[n+n-1+n-2+n-3]=178\Rightarrow 4n-6=50\Rightarrow n=14\therefore \text{Median}=\frac{t_7+t_8}{2}=\frac{14+3\times 15}{2}=29.5$