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The sum of first 40 positive integers divisible by 6 is (a) 2460 (b) 3640 (c) 4920 (d) 4860 |
Answer» Correct answer is (c) 4920 The positive integers divisible by 6 are 6, 12, 18,.... This is an AP with a = 6 and d = 6. Also, n = 40 (Given) Using the formula, \(S_n=\frac{n}{2}[2a+(n-1)d],\) we get \(S_{40}=\frac{40}{2}[2\times6+(40-1)\times6]\) = 20(12 + 234) = 20 x 246 = 4920 Thus, the required sum is 4920 |
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