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Question
How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116 ? Also find the last term.
Solution
A.P. is 25, 22, 19, …
Sum = 116
Here, a = 25, d = 22 – 25 = -3
Let number of terms be n, then
116 = `n/(2)[2a + (n - 1)d]`
⇒ 232 = n[2 x 25 + (n – 1)(– 3)]
⇒ 232 = n[50 – 3n + 3]
⇒ n(53 – 3n)
⇒ 232 = 53n – 3n2
⇒ 3n2 – 53n + 232 = 0 ...`{(∵232 xx 3, = 696),(∴ 696, = -24 xx (-29)),(-53, = -24 - 29):}}`
⇒ 3n2 – 24n – 29n + 232 = 0
⇒ 3n(n – 8) – 29(n – 8) = 0
⇒ (n – 8)(3n – 29) = 0
Either n – 8 = 0,
then n = 8
or
3n – 29 = 0,
then 3n = 29
⇒ n = `(29)/(3)`
which is not possible because of fraction
∴ n = 8
Now, T = a + (n – 1)d
= 25 + 7 x (–3)
= 25 – 21
= 4.
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