Question

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

Answer 1

Answer:

First 40 +ve integers which are divisible by 6

= 6,12,18,24..........240

Here ,

a = 6

d = 6

n = 40

Total sum = n/2 { 2a + (n - 1)d}

= 40/2{ 2×6 + (40 - 1)6

= 20{ 12 + 234 }

= 20 × 246

= 4920 Ans...

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Answer 2

Sn = n/2×{2a+(n-1)d}

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6 , 12 18 .... are numbers divisible by 6

a = 6

d = 6

n =40

substituting in the formula

n/2×{2a+(n-1)d}

s = 40/2 x {(2(6) + (40-1)6)}

20 ( 12 + 234 )

20(246)

=4920

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