Find the third term of a geometric progression if the sum of the first three terms is equal to 12, and the sum of the first six terms is equ

Question

Find the third term of a geometric progression if the sum of the first three terms is equal to 12, and the sum of the first six terms is equal to (−84).

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Dâu 2 months 2021-07-24T22:48:06+00:00 1 Answers 4 views 0

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    2021-07-24T22:49:45+00:00

    Given:

    The sum of the first three terms = 12

    The sum of the first six terms = (−84).

    To find:

    The third term of a geometric progression.

    Solution:

    The sum of first n term of a geometric progression is:

    S_n=\dfrac{a(r^n-1)}{r-1}

    Where, a is the first term and r is the common ratio.

    The sum of the first three terms is equal to 12, and the sum of the first six terms is equal to (−84).

    \dfrac{a(r^3-1)}{r-1}=12               …(i)

    \dfrac{a(r^6-1)}{r-1}=-84               …(ii)

    Divide (ii) by (i), we get

    \dfrac{r^6-1}{r^3-1}=\dfrac{-84}{12}

    \dfrac{(r^3-1)(r^3+1)}{r^3-1}=-7

    r^3+1=-7

    r^3=-7-1

    r^3=-8

    Taking cube root on both sides, we get

    r=-2

    Putting r=-2 in (i), we get

    \dfrac{a((-2)^3-1)}{(-2)-1}=12

    \dfrac{a(-8-1)}{-3}=12

    \dfrac{-9a}{-3}=12

    3a=12

    Divide both sides by 3.

    a=4

    The nth term of a geometric progression is:

    a_n=ar^{n-1}

    Where, a is the first term and r is the common ratio.

    Putting n=3,a=4,r=-2 in the above formula, we get

    a_3=4(-2)^{3-1}

    a_3=4(-2)^{2}

    a_3=4(4)

    a_3=16

    Therefore, the third term of the geometric progression is 16.

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