Let 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 _________ . [2023]

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Solution

Q.
Let $a_{1} = 8, a_{2}, a_{3}, . . ., a_{n}$ 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 _________ . [2023]

Ans.
(754)

Sum of first four terms $= \frac{4}{2} [16 + 3 d]$

$\Rightarrow 50 = 32 + 6 d \Rightarrow d = 3$

Now, sum of last four terms = 170

$\Rightarrow \frac{4}{2} [2 (8 + (n - 4) d) + 3 d] = 170$ $[∵ last four terms are a_{n - 3}, a_{n - 2}, a_{n - 1}, a_{n}]$

$\Rightarrow 2 [2 (8 + (n - 4) \cdot 3) + 9] = 170$

$\Rightarrow 16 + 6 (n - 4) + 9 = 85$

$\Rightarrow 6 (n - 4) = 60 \Rightarrow n - 4 = 10 \Rightarrow n = 14$

$∴$ $Middle terms are 7th term and 8th term.$

$∴$ $T_{7} = 8 + 6 d = 8 + 18 = 26, T_{8} = 8 + 7 d = 8 + 21 = 29$

Now, $T_{7} \times T_{8} = 26 \times 29 = 754$

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