If $a_{1}$ , $a_{2}$ , $a_{3}$ , .... is an A.P. such that $a_{1}$ + $a_{5}$ + $a_{10}$ + $a_{15}$ + $a_{20}$ + $a_{24}$ = 225 then $a_{1}$ + $a_{2}$ + $a_{3}$ + ... + $a_{23}$ + $a_{24}$ is equal to

909

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750

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900

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Solution

The correct option is D
900

We have, $a_{5}$ + $a_{20}$ = $a_{1}$ + $a_{24}$ , $a_{10}$ + $a_{15}$ = $a_{1}$ + $a_{24}$

Hence the given relations reduce to, 3( $a_{1}$ + $a_{24}$ ) = 225, giving $a_{1}$ + $a_{24}$ = 75

Hence $S_{24}$ = $\frac{n}{2}$ (a + l) = ( $\frac{24}{2}$ )( $a_{1}$ + $a_{24}$ ) = 12 × 75 = 900

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Similar questions

a_{1}, a_{2}, a_{3}, . . . .

a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225

a_{1} + a_{2} + a_{3} + . . . + a_{23} + a_{24}

If $a_{1}, a_{2}, a_{3} . . . . .$ are in $A . P .$ such that $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$ .

a_{1} + a_{2} + a_{3} . . . . . . . . . + a_{23} + a_{24}

a_{1}, a_{2}, a_{3}, . . . .

a_{1}, a_{5}, a_{10} + a_{15} + a_{20} + a_{24} = 225

a_{1} + a_{2} + a_{3} + . . . + a_{24}

a_{1}, a_{2}, a_{3}, \dots

a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225

a_{1} + a_{2} + a_{3} + \dots a_{23} + a_{24}

a_{1}, a_{2}, a_{3}, . . . . . . . . i s a n A . P . s u c h t h a t a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225

a_{1} + a_{2} + a_{3} + . . . . . . . + a_{23} + a_{24} i s e q u a l t o

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