ow many terms of the ap 1 4 7 are needed to make the sum 51

Question

ow many terms of the ap 1 4 7 are needed to make the sum 51

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Melanie 4 weeks 2021-08-17T15:34:44+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-17T15:36:29+00:00

    Answer :

    \large{\star\:\:\boxed{\sf{S\sf{_6}\:=\:51}}\:\:\star}

    Explanation :

    Given :–

    • 1 , 4 , 7 , 10 , … are in A.P.
    • where a = 1 and d = 3 .

    To Find :–

    • How many terms(n) will give a Sum of 51 (Sₙ)  .

    Formula Applied :–

    • \boxed{\star\:\:\bf{S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d] }\:\:\star}

    Solution :–

    We have ,

    • a = 1
    • d = 3
    • Sₙ = 51

    Putting these values in the Formula :

    \rightarrow\sf{S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d] }

    \rightarrow\sf{51\:=\:\dfrac{n}{2}\:[2(1)\:+\:(n\:-\:1)(3)] }

    \rightarrow\sf{51\:\times\:2\:=\:n\:(2\:+\:3n\:-\:3) }

    \rightarrow\sf{102\:=\:n\:(3n\:-\:1) }

    \rightarrow\sf{102\:=\:3n^2\:-\:n }

    \rightarrow\sf{3n^2\:-\:n\:-\:102\:=\:0 }

    \rightarrow\sf{3n^2\:-\:18n\:+\:17n\:-\:102\:=\:0 }

    \rightarrow\sf{3n(n\:-\:6)\:+\:17(n\:-\:6)\:=\:0 }

    \rightarrow\sf{(3n\:+\:17)(n\:-\:6)\:=\:0 }

    \rightarrow\sf{n\:=\:6\:,\:(-\dfrac{17}{3})  }

    Since the Number of terms (n) can’t be negative !

    So , the Positive value will be correct , i.e.

    \rightarrow\boxed{\bf{n\:=\:6 }}

    6 Terms of this A.P. will give a Sum of 51 .

    0
    2021-08-17T15:36:29+00:00

    Answer:

    first term (a)=1

    common difference (d)=3

    we know

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

    =>51=n/2 {2.1+(n-1) 3}=n/2 {3n-1}

    =>3n^2-n=102

    => 3n^2-n-102=0

    use quadratic formula

    n=6,-17/3

    but n =-17/3 is not possible

    so, n=6

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