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Question
How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78? Explain the double answer.
Solution
A.P. is 24, 21, 18,...
Sum = 78
Here, a = 24, d = 21 – 24 = –3
Sn = `n/(2)[2a + (n – 1)d]`
⇒ 78 = `n/(2)[2 xx 24 + (n - 1)(-3)]`
⇒ 156 = n(48 – 3n + 3)
⇒ 156 = 51n – 3n2
⇒ 3n2 – 51n + 156 = 0
⇒ 3n2 – 12n – 39n + 156 = 0 ...`{(∵ 156 xx 3, = 468),(∴ 468, = -12 x - 39),(-51, = -12 - 39):}}`
⇒ 3n(n – 4) – 39(n – 4) = 0
⇒ (n – 4)(3n – 39) = 0
Either n – 4 = 0,
then n = 4
or
3n – 39 = 0,
then 3n = 39
⇒ n = 13
∴ n = 4 and 13
n4 = a + (n – 1)d
= 24 + 3(–3)
= 24 – 9
= 15
n13 = 24 + 12(–3)
= 24 – 36
= –12
∴ Sum of 5th term to 13 term = 0
(∵ 12 + 9 + 6 + 3 + 0 + (–3) + (–6) + (–9) + (–12) = 0.
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