Find the sum of the first 40 positive integers divisible by 6

1.	Find the sum of the first 40 positive integers divisible by 6
Answer» The first 40 positive integers divisible by 6 are 6, 12, 18, 24, .....Here, a2 - a1 = 12 - 6 = 6a3 - a2 = 18 - 12 = 6a4 - a3 = 24 - 18 = 6i.e. ak+1 - ak is the same everytime.So, the above list of numbers from an AP.Here, a = 6d = 6n = 40{tex}\\therefore {/tex}\xa0Sum of the first 40 positive integers = S40{tex} = \\frac{{40}}{2}\\left[ {2a + (40 - 1)d} \\right]{/tex}\xa0.........{tex}\\because {S_n} = \\frac{n}{2}\\left[ {2a + (n - 1)d} \\right]{/tex}= 20[2a + 39d]{tex} = (20)[2 \\times 6 + 39 \\times 6]{/tex}= (20) (246)= 4920

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