The sum of first $$26$$ terms of an A.P $$a1$$,$$a2$$,$$a3$$,… , If $$a2+a6+a9+a18+a21+a25=165$$, is

We $$havea1+a26=a2+a25=a3+a24=$$…$$=a9+a18Thus$$, $$a2+a6+a9+a18+a21+a25=165$$⇒$$a2+a25+a6+a21+a9+a18=165$$⇒$$3a1+a26=165$$⇒$$a1+a26=55$$∴$$a1+a2+$$…$$+a26=262a1+a26=13$$×$$55=715$$

Arithmetic progression

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Question

The sum of first $26$ terms of an A.P $a_{1}, a_{2}, a_{3}, \dots,$ If $a_{2} + a_{6} + a_{9} + a_{18} + a_{21} + a_{25} = 165,$ is

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Solution

We have
$a_{1} + a_{26} = a_{2} + a_{25} = a_{3} + a_{24} = \dots = a_{9} + a_{18}$
Thus,
$\begin{array}{l} a_{2} + a_{6} + a_{9} + a_{18} + a_{21} + a_{25} = 165 \\ \Rightarrow (a_{2} + a_{25}) + (a_{6} + a_{21}) + (a_{9} + a_{18}) = 165 \\ \Rightarrow 3 (a_{1} + a_{26}) = 165 \Rightarrow a_{1} + a_{26} = 55 \\ ∴ a_{1} + a_{2} + \dots + a_{26} = \frac{26}{2} (a_{1} + a_{26}) \\ = 13 \times 55 = 715 \end{array}$

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Arithmetic progression

Remember concepts with our Masterclasses.

The sum of first 26 terms of an A.P a1,a2,a3,… , If a2+a6+a9+a18+a21+a25=165, is

We havea1+a26=a2+a25=a3+a24=…=a9+a18Thus, a2+a6+a9+a18+a21+a25=165⇒a2+a25+a6+a21+a9+a18=165⇒3a1+a26=165⇒a1+a26=55∴a1+a2+…+a26=262a1+a26=13×55=715

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The sum of first $26$ terms of an A.P $a_{1}, a_{2}, a_{3}, \dots,$ If $a_{2} + a_{6} + a_{9} + a_{18} + a_{21} + a_{25} = 165,$ is