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Question
Find the sum of first 24 terms of the A.P. a1, a2, a3, ... if it is known that a1 + a5 + a10 + a15 + a20 + a24 = 225.
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Solution
We know that in an A.P., the sum of the terms equidistant from the beginning and end is always the same and is equal to the sum of first and last term.
Therefore d = b – a
i.e., a1 + a24 = a5 + a20
= a10 + a15
It is given that (a1 + a24) + (a5 + a20) + (a10 + a15) = 225
⇒ (a1 + a24) + (a1 + a24) + (a1 + a24) =225
⇒ 3(a1 + a24) = 225
⇒ a1 + a24 = 75
We know that Sn = `n/2[a + 1]`
Where a is the first term and l is the last term of an A.P.
Thus, S24 = `24/2 [a_1 + a_24]`
= 12 × 75
= 900
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