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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

Option: 1

754


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \mathrm{a}_1=8 \\ & \mathrm{~d}=\text { common difference } \\ & \frac{4}{2}[16+3 \mathrm{~d}]=50 \\ & \Rightarrow \mathrm{d}=3 \end{aligned}

$$ \begin{gathered} \frac{4}{2}\left[2 a_n+3(-d)\right]=170 \\ \Rightarrow 2\left(a_1+(n-1) d\right)-3 d=85 \\ \Rightarrow 16+6(n-1)-9=85 \\ \quad n-1=13 \\ \quad n=14 \end{gathered}
Product of middle two terms =\mathrm{T}_7 \times \mathrm{T}_8
$$ \begin{aligned} & =\left(\mathrm{a}_1+6 \mathrm{~d}\right)\left(\mathrm{a}_1+7 \mathrm{~d}\right) \\ & =(8+18)(8+21) \\ & =(26)(29)=754 \end{aligned}

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Rakesh

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