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Question
In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P. must be taken for their sum to be equal to 120?
Options
6
7
8
9
Solution
8
Explanation;
Here a = 1, d = 4, Sn = 120
Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`
120 = `"n"/2 [2 + ("n" - 1)4]`
= `"n"/2 [2 + 4"n" - 4]`
= `"n"/2 [4"n" - 2]`
= `"n"/2 xx 2 (2"n" - 1)`
120 = 2n2 – n
∴ 2n2 – n – 120 = 0
⇒ 2n2 – 16n + 15n – 120 = 0
2n(n – 8) + 15 (n – 8) = 0
⇒ (n – 8) (2n + 15) = 0
n = 8 or n = `(-15)/2` ...(omitted)
∴ n = 8
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