If $a_{1}, a_{2}, a_{3}, . . . . a_{24}$ are in arithmetic progression and $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225,$ then $a_{1} + a_{2} + a_{3} + . . . . + a_{23} + a_{24} =$

909

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750

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900

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Solution

The correct option is D 900
$a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$
$\Rightarrow (a_{1} + a_{24}) + (a_{5} + a_{20}) + (a_{10} + a_{15}) = 225$
$\Rightarrow 3 (a_{1} + a_{24}) = 225 \Rightarrow a_{1} + a_{24} = 75$
$∵ In an A.P. the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of first and last term$
$a_{1} + a_{2} + . . . + a_{24} = \frac{24}{2} (a_{1} + a_{24}) = 12 \times 75 = 900.$

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