Let a_1=8, a_2, a_3, , a_n be an A.P.

Question

Let $a_1=8, a_2, a_3, \ldots, a_n$ be an A.P. If the sum of its first four terms is $50$ and the sum of its last four terms is $170$, then the product of its middle two terms is \_\_\_\_\_\_\_\_.

Answer: 754

Accepted Answer

$a_1+a_2+a_3+a_4 = 32+6d = 50 \Rightarrow d=3$.

$a_{n-3}+a_{n-2}+a_{n-1}+a_n = 32+(4n-10)(3) = 170 \Rightarrow n=14.$

The middle two terms are $a_7=26$ and $a_8=29$.

$$a_7\cdot a_8 = 26\times29 = 754.$$

Solutions · 1