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 find the value of

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

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Solution

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

⇒( $a_{1}$ + $a_{24}$ ) + ( $a_{5}$ + $a_{20}$ ) + ( $a_{10}$ + $a_{15}$ ) = 225

⇒ 3( $a_{1}$ + $a_{24}$ ) = 225 ⇒ $a_{1}$ + $a_{24}$ = 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 × 75 = 900.

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

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A . P .

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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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