In question of class 10th of arithmetic progressions of exercise 5.3, this is the best question for class 10th for board exams Find the sum of first 40 positive integers divisible by 6.
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The positive integers that are divisible by 6 are 6, 12, 18, 24 ….
We can see here, that this series forms an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
S40 = ?
By the formula of sum of n terms, we know,
Sn = n/2 [2a +(n – 1)d]
Therefore, putting n = 40, we get,
S40 = 40/2 [2(6)+(40-1)6]
= 20[12+(39)(6)]
= 20(12+234)
= 20×246
= 4920