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

To solve for the product of the middle two terms of the given arithmetic progression (AP), follow these steps:

Step 1: Understand the AP Parameters 
Given an AP: a₁=8, a₂, a₃, ..., a_n. The sum of the first four terms is 50, and the sum of the last four terms is 170.

Step 2: Find the Common Difference
The first four terms are: a₁=8, a₂, a₃, a₄. The sum is:
 

8 + (8 + d) + (8 + 2d) + (8 + 3d) = 50

32 + 6d = 50

6d = 18

d = 3

Step 3: Determine n
The last four terms are: a_n-3, a_n-2, a_n-1, a_n, with sum:
 

(8 + (n-4)d) + (8 + (n-3)d) + (8 + (n-2)d) + (8 + (n-1)d) = 170

4(8) + (4n - 10)d = 170

32 + 12n - 30 = 170

12n = 168

n = 14

Step 4: Calculate Middle Terms and Their Product
With n = 14, the AP terms are: a₁ = 8, a₂=11, ..., a₁₄. Middle terms: a₇, a₈.
 

a₇ = 8 + 6d = 8 + 18 = 26

a₈ = 8 + 7d = 8 + 21 = 29

The product: 26 × 29 = 754

Step 5: Validate Solution
The product of the middle terms, 754, falls within the specified range (754,754). Thus, this is the correct and validated solution.

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: \[ a_7 = 26, \quad a_8 = 29 \] \[ \Rightarrow a_7 \cdot a_8 = 754 \]

Answer 3

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