Question

An AP consists of $23$terms. If the sum of the three terms in the middle is $141.$ and the sum of the last three terms is$261$, then the first term is

A

$6$

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B

$5$

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C

$4$

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D

$3$

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Solution

The correct option is D $3$The explanation for the correct option:Step 1. Assume the variables :Let ${a}_{1},{a}_{2}.........,{a}_{23}$be the terms in AP.Let $d$be the common difference.Step 2. Apply the given condition :the sum of the three terms in the middle is $141.$$⇒{a}_{11}+{a}_{12}+{a}_{13}=141\phantom{\rule{0ex}{0ex}}⇒a+10d+a+11d+a+12d=141\mathbf{\left[}\mathbf{\because }\mathbf{}{\mathbf{a}}_{\mathbf{n}}\mathbf{=}\mathbf{}\mathbf{a}\mathbf{}\mathbf{+}\mathbf{\left(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{\right)}\mathbf{d}\mathbf{\right]}\phantom{\rule{0ex}{0ex}}⇒3a+33d=141\phantom{\rule{0ex}{0ex}}⇒a+11d=47..............\left(1\right)$the sum of the last three terms is $261.$$⇒{a}_{21}+{a}_{22}+{a}_{23}=261\phantom{\rule{0ex}{0ex}}⇒a+20d+a+21d+a+22d=261\phantom{\rule{0ex}{0ex}}⇒3a+63d=261\phantom{\rule{0ex}{0ex}}⇒a+21d=87.............\left(2\right)$Step3. Solving the equation :Solving $\left(1\right)$ and $\left(2\right)$, we get $d=4$Substitute $d$in $\left(1\right)$, we get$⇒a+44=47\phantom{\rule{0ex}{0ex}}⇒a=3$Thus, the first term is $3$.Hence, the correct option is D.

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