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How many terms of the AP : 24, 21, 18, ... must be taken so that their sum is 78? |
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Answer» Given A.P. 24, 21, 18,..... First term = 24 = a and common difference = – 3 = d ...(i) Let no. of terms is n. Sum of n terms = \(\frac{n}{2} [2a + (n - 1)d]\) According to question \(78 = \frac{n}{2} [2 \times 24 - 3 \times (n -1)]\) [from (i) and given] \(78 = \frac{n}{2} [51- 3n]\) n2 – 17n + 52 = 0 n2 – 13n – 4n + 52 = 0 n(n – 13) – 4 (n – 13) = 0 (n – 13) (n – 4) = 0 n = 13, 4 For first 4 terms and first 13 terms in both case we get sum 78. |
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