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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75

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750

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900

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Solution

The correct option is D $900$
Here we use the following formula to calculate the sum of first n terms of an AP:
$S_{n} = \frac{n}{2} (2 a + (n - 1) d)$
Since $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$
$\Rightarrow 6 a + 69 d = 225$
$\Rightarrow 2 a + 23 d = 75$
Now $S_{24} = a_{1} + a_{2} + a_{3} + . . . + a_{23} + a_{24}$
$S_{24} = 12 (2 a + 23 d)$
$= 900$

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Q. 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

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

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

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

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

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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

then

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

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If a1,a2,a3,.... is 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 900Here we use the following formula to calculate the sum of first n terms of an AP: Sn=n2(2a+(n−1)d)Since a1+a5+a10+a15+a20+a24=225⇒6a+69d=225⇒2a+23d=75Now S24=a1+a2+a3+...+a23+a24 S24=12(2a+23d) =900

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-