Let the sum of the first three terms of an A.P.

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:

Accepted Answer

$a_1+a_2+a_3=39$, with $a_1=10$: $3(10)+3d=39 \implies d=3$

Mean of last four terms $= \dfrac{178}{4}=44.5$; since consecutive terms average to the middle, $a_n = 44.5+1.5+3=49$ *(using the recovered relation for the last-four-terms structure)*.

$$\text{Median} = \frac{10+49}{2}=\frac{59}{2}=29.5$$

Solutions · 1