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

600

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900

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1200

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1800

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Solution

The correct option is B 900
$a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$ (given)
$a + (a + 4 d) + (a + 9 d) + (a + 14 d) + (a + 19 d) + (a + 23 d) = 225$
$6 a + 69 d = 225 . . . (i)$
$a_{1}, a_{2}, a_{3} . . .$ in A.P
$a_{1} a_{2} = a + d, a_{3} = a + 2 d$ etc
$a_{1} + a_{2} + a_{3} + . . . . + a_{24} = ?$
$n = 24, S_{24} = \frac{n}{2} [2 a + (n - 1) d] = \frac{24}{2} [2 a + 23 d]$
$= 12 [75]$ [from eq. (i)]
= 900 Ans.
from eq. (i)
$3 [2 a + 23 d] = 225$
$∴ 2 a + 23 d = 75... (i i)$

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