If $a_{1}, a_{2}, a_{3}, \dots$ are an A.P such that $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$ , then $a_{1} + a_{2} + a_{3} + \dots a_{23} + a_{24}$ 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
$a_{10} + a_{15} = a_{1} + a_{24}, a_{5} + a_{20} = a_{1} + a_{24}$
Hence the given relations reduce to
$3 (a_{1} + a_{24}) = 225$ , giving $a_{1} + a_{24} = 75$ ,
Hence $s_{24} = \frac{n}{2} (a + l) = (24 / 2) (a_{1} + a_{24}) = 12 \times 75 = 900$

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If a1,a2,a3,⋯ are an A.P such that a1+a5+a10+a15+a20+a24=225, then a1+a2+a3+⋯a23+a24 is equal to

The correct option is D 900a10+a15=a1+a24,a5+a20=a1+a24 Hence the given relations reduce to 3(a1+a24)=225, giving a1+a24=75, Hence s24=n2(a+l)=(24/2)(a1+a24)=12×75=900

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

The correct option is D 900
$a_{10} + a_{15} = a_{1} + a_{24}, a_{5} + a_{20} = a_{1} + a_{24}$
Hence the given relations reduce to
$3 (a_{1} + a_{24}) = 225$ , giving $a_{1} + a_{24} = 75$ ,
Hence $s_{24} = \frac{n}{2} (a + l) = (24 / 2) (a_{1} + a_{24}) = 12 \times 75 = 900$