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Question
Find the sum of first 40 positive integers divisible by 6.
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Solution 1
The positive integers divisible by 6 are:
6, 12, 18, 24, .......
a = 6, d = 12 - 6 = 6, n = 40
∵ Sn = `"n"/2` [2a + (n - 1) × d]
⇒ S40 = `40/2` [2 × 6 + (40 - 1) × 6]
20[12 + 39 × 6]
= 20[12 + 234]
= 20(246)
= 4920
Solution 2
The positive integers that are divisible by 6 are
6, 12, 18, 24...
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
S40 =?
The positive integers that are divisible by 6 are
6, 12, 18, 24...
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
S40 =?
`S_n = n/2 [2a+(n-1)d]`
`S_40 = 40/2 [2(6)+(40-1)6]`
= 20 [12 + (39) (6)]
= 20 (12 + 234)
= 20 × 246
= 4920
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