find the sum of the first 30 positive integer divisible by 6

Question

find the sum of the first 30 positive integer divisible by 6

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Remi 5 months 2021-07-08T01:29:24+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-08T01:30:25+00:00

    Step-by-step explanation:

    6,12,18,24,30

    a=6

    d=6

    n=30

    Sn=?

    Sn=n/2(2a+(n-1)d

    30/2(2×6+(30-1)6

    15(12+(29)6

    15(12+174)

    15(186)

    2790

    0
    2021-07-08T01:31:23+00:00

    Step-by-step explanation:

    Given:

    First 30 positive integers divisible by 6

    To find:

    Find the sum of the first 30 positive integers which are divisible by 6 ?

    Solution:

    The list of positive integers

    = 1,2,3,4,…

    The list of positive integers which are divisible by 6

    = 6,12,18,24,30,…

    The list of first 30 positive integers which are divisible by 6

    = 6,12,18,…(30 terms)

    First term = a = 6

    Common difference = 12-6=6

    Since the common difference is same throughout the series

    6,12,18,…are in the AP

    Now we have to find the sum of first 30 positive integers which are divisible by 6

    = 6+12+18+…+(30 terms)

    We know that

    The sum of first n terms in an AP

    = Sn = (n/2)[2a+(n-1)d]

    We have

    a = 6

    d = 6

    n = 30

    Now,

    S 30 = (30/2)[2(6)+(30-1)(6)]

    => S 30 = (15)[12+29(6)]

    => S 30 = (15)(12+174)

    => S 30 = 15(186)

    => S 30 = 2790

    Answer:

    The sum of first 30 positive integers which are divisible by 6 is 2790

    Used formulae:

    The sum of first n terms in an AP

    Sn = (n/2)[2a+(n-1)d]

    • n = number of terms
    • a = First term
    • d = Common difference

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