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Sum of n Terms

If a1,a2,a3...

Question

If $a_{1}, a_{2}, a_{3}, . . . . . . . . i s a n A . P . s u c h t h a t a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$
then $a_{1} + a_{2} + a_{3} + . . . . . . . + a_{23} + a_{24} i s e q u a l t o$

909

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75

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750

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900

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Solution

The correct option is D $900$
$a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$
$a_{1} + (a_{1} + 4 d) + (a_{1} + 9 d) + (a_{1} + 14 d)$
$(a_{1} + 19 d) + (a_{1} + 23 d) = 225$
$6 a_{1} + 69 d = 225$
$2 a_{1} + 23 d = \frac{225}{3}$
$N o w a_{1} + . . . . . . . . + a_{24}$
$= \frac{24}{2} [2 a_{1} + (2 a - 1) d]$
$= 12 [2 a_{1} + 23 d]$
$= 12 \times \frac{225}{3}$
$= 4 \times 225$
$= 900$

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Similar questions

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} =

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

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

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}, \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}

A . P .

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

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

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