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

909

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4 \times 225

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750

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990

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Solution

The correct option is B $4 \times 225$
$a + a + 4 d + a + 9 d + a + 14 d + a + 19 d + a + 23 d = 6 a + 69 d = 225,$
$3 (2 a + 23 d) = 225$
$2 a + 23 d = \frac{225}{3}$
$S_{n} = \frac{n}{2} [2 a + (n - 1) d], S_{n} = \frac{24}{2} [2 a + 23 d]$
$S_{n} = 12 [2 a + 23 d], S_{n} = 12 * 225 / 3 \Rightarrow S_{n} = 900$

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