Question

Let $a_1=8, a_2, a_3, \ldots, a_n$ be an AP 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 1

Correct Answer: 754

The correct answer is 754.
$a_1+a_2+a_3+a_4=50$
$\Rightarrow 32+6d=50$
$\Rightarrow d=3$
$\text{and,}{a}_{n-3}+a_{n-2}+a_{n-1}+a_n=170$
$\Rightarrow 32+(4n-10)\cdot 3=170$
$\Rightarrow n=14$
$a_7=26,a_8=29$
$\Rightarrow a_7\cdot a_8=754$

Answer 2

Using the sum formula for A.P., \[ a_1 + a_2 + a_3 + a_4 = 50 \] \[ 8 + (8 + d) + (8 + 2d) + (8 + 3d) = 50 \] \[ 32 + 6d = 50 \Rightarrow d = 3 \] For the last four terms, \[ a_{n-3} + a_{n-2} + a_{n-1} + a_n = 170 \] \[ 32 + (4n - 10) \cdot 3 = 170 \] \[ n = 14 \] Middle terms are: \[ a_7 = 26, \quad a_8 = 29 \] \[ \Rightarrow a_7 \cdot a_8 = 754 \]