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Question
How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693? Explain the double answer.
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Solution
The given AP is 63, 60, 57, 54, ...
Here, a = 63 and d = 60 – 63 = –3
Let the required number of terms be n. Then,
Sn = 693
⇒ `n/2 [2 xx 63 + (n - 1) xx (-3)] = 693` ...`{S_n = n/2 [2a + (n - 1)d]}`
⇒ `n/2 (126 - 3n + 3) = 693`
⇒ n(129 – 3n) = 1386
⇒ 3n2 – 129n + 1386 = 0
⇒ 3n2 – 66n – 63n + 1386 = 0
⇒ 3n(n – 22) – 63(n – 22) = 0
⇒ (n – 22) (3n – 63) = 0
⇒ n – 22 = 0 or 3n – 63 = 0
⇒ n = 22 or n = 21
So, the sum of 21 terms as well as that of 22 terms is 693. This is because the 22nd term of the AP is 0.
a22 = 63 + (22 – 1) × (–3)
= 63 – 63
= 0
Hence, the required number of terms is 21 or 22.
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