Math, asked by vaishnavi7210, 1 year ago

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

Answers

Answered by drmalik021
22

hi friend,


the first 40 positive integers divisible by 6 are 6,12,18,....... upto

40 terms


the given series is in arthimetic progression with first term a=6 and common difference d=6

sum of n terms of an A.p is

n/2×{2a+(n-1)d}


→required sum = 40/2×{2(6)+(39)6}

=20{12+234}

=20×246

=4920


I hope this will help u ;)



Answered by muskan2807
2

Answer:

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 and d = 6

sn=n/2(2a+(n-I) d)

S40 =?

= 20[12 + (39) (6)]

= 20(12 + 234)

= 20 × 246

= 4920

please mark my answer as brainlist.

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