If $a_{1}, a_{2}, a_{3}, . . . . . . . . .$ be in $A . P .$ such that $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$ , then the sum of first $24$ terms of the $A . P$ is

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Solution

Given, $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$
$\Rightarrow a + (a + 4 d) + (a + 9 d) + (a + 14 d) + (a + 19 d) + (a + 23 d) = 225$
$\Rightarrow 6 a + 69 d = 225$
$\Rightarrow 2 a + 23 d = 75$ ---(1)
Now, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$S_{24} = \frac{24}{2} [2 a + (24 - 1) d]$
$= 12 \times (2 a + 23 d)$
$= 12 \times 75$ ...... (from (1))
$= 900$

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