Find the sum of the first 40 positive integers divisible by 6.
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3
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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Answered by
1
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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