how many terms of the AP:9, 17, 25,… must be taken to give a sum of 636​

Question

how many terms of the AP:9, 17, 25,… must be taken to give a sum of 636​

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Madelyn 2 weeks 2021-09-09T13:06:07+00:00 2 Answers 0 views 0

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    0
    2021-09-09T13:07:34+00:00

    \bf{\underline{Question:-}}

    how many terms of the AP:9, 17, 25,… must be taken to give a sum of 636.

    \bf{\underline{Given:-}}

    • First term (a) = 9
    • Common difference (d) = a2 – a1 = 17 – 9= 8
    • Sum (Sn)= 636

    \bf{\underline{Formula:-}}

    \bf\large S_n= \frac{n}{2}[(2a+(n-1)d]

    \bf{\underline{Solution:-}}

    Substituting values in formula

    \bf →  S_n = \frac{n}{2}[2a+(n-1)d]

    \bf →636 = \frac{n}{2}[2(9)+(n-1)8

    \bf →636=\frac{n}{2}[18 + 8n - 8]

    \bf →636 = \frac{n}{2}[8n + 10]

    \bf →4n^2 + 5n - 636 = 0

    \bf →4n^2 + (53-48)n - 636=0

    \bf → 4n^2 + 53n - 48n - 636 = 0

    \bf →4n ( n - 12) + 53(n-12)=0

    \bf →(4n + 53 ) ( n - 12 ) = 0

    \bf →4n + 53 = 0

    \bf →4n = -53

    \bf → n = \frac{-53}{4}

    Either

    \bf →n - 12 = 0

    \bf →n= 12

    \bf{\underline{Therefore:-}}

    • – 53/4 is negative So, we neglect it

    \bf{\underline{Hence:-}}

    • 12 term are needed to give the sum of an A.P 636
    0
    2021-09-09T13:08:03+00:00

    Answer:

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