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Find the sum of the first 40 positive integers divisible by 6.
Find the sum of the first 40 positive integers divisible by 6.
Updated On: Jan 13, 2026
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Solution and Explanation
The sequence of positive integers divisible by 6 is \(6, 12, 18, 24, \dots\). This forms an arithmetic progression (A.P.) with the first term \(a = 6\) and a common difference \(d = 6\). We need to find the sum of the first 40 terms, \(S_{40}\). The formula for the sum of an A.P. is \(S_n = \frac n2 [2a + (n-1)d]\).
Substituting the values, we get \(S_{40} = \frac {40}{2} [2(6) + (40-1)6]\). This simplifies to \(S_{40} = 20[12 + (39) (6)]\), then \(S_{40} = 20(12 + 234)\), and further to \(S_{40} = 20 \times 246\). Therefore, \(S_{40 }= 4920 \).
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