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Question
How many terms of the A.P. : 24, 21, 18, ................ must be taken so that their sum is 78?
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Solution 1
Let the number of terms taken be n.
The given A.P. is 24, 21, 18, ................
Here, a = 24 and d = 21 – 24 = –3
`S = n/2 [2a + (n - 1)d]`
`=> 78 = n/2 [2 xx 24 + (n - 1)xx (-3)]`
`=> 78 = n/2 [48 - 3n + 3]`
`=>` 156 = n[51 – 3n]
`=>` 156 = 51n – 3n2
`=>` 3n2 – 51n + 156 = 0
`=>` n2 – 17n + 52 = 0
`=>` n2 – 13n – 4n + 52 = 0
`=>` n(n – 13) – 4(n – 13) = 0
`=>` (n – 13)(n – 4) = 0
`=>` n = 13 or n = 4
∴ Required number of terms = 4 or 13
Solution 2
Given: 24, 21, 18, ........ are in A.P.
Here, a = 24, d = 21 – 24 = – 3
Sum, `S_n = n/2 [2a + (n - 1)d]`
⇒ `78 = n/2 [2 xx 24 + (n - 1)(-3)]`
⇒ 156 = n(48 – 3n + 3)
⇒ 156 = n(51 – 3n)
⇒ 3n2 – 51n + 156 = 0
⇒ 3n2 – 12n – 39n + 156 = 0
⇒ 3n(n – 4) – 39(n – 4) = 0
⇒ (n – 4)(3n – 39) = 0
∴ (n – 4) = 0 and (3n – 39) = 0
⇒ n = 4 and n = 13
When n = 4,
`S_4 = 4/2 [2 xx 24 + (4 - 1)(-3)]`
= 2(48 – 9)
= 2 × 39
= 78
When n = 13
`S_13 = 13/2 [2 xx 24 + (13 - 1)(-3)]`
= `13/2 [48 + (-36)]`
= `13/2 xx 12`
= 78
Hence, the number of terms is n = 4 or n = 13.
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