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

909

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750

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900

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Solution

The correct option is D 900
Given, $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225 \Rightarrow (a_{1} + a_{24}) + (a_{5} + a_{20}) + (a_{10} + a_{15}) = 225 \Rightarrow 3 (a_{1} + a_{24} = 225) \Rightarrow a_{1} + a_{24} = \frac{225}{3} ∴ a_{1} + a_{2} + a_{3} + . . . . + a_{24} (a_{1} + a_{24}) = \frac{24}{2} (\frac{225}{3}) = 900$

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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_{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 . P .

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}

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}

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