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In an AP, Of Sn=n(4n+1) , …

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In an AP, Of Sn=n(4n+1) , find the Ap

    • Posted by Rutuja Londhe 1 year, 4 months ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    S P Jaishwal 1 year, 4 months ago

    Sn = n(4n+1) Sn = 4n^2 +n Sn-1 = 4(n-1)^2 +(n-1) Sn-1 = 4(n^2+1-2n) +(n-1) Sn-1 = 4n^2+4-8n +n -1 Sn-1 = 4n^2-7n -3 Sn - Sn-1 = 4n^2+ n -(4n^2-7n-3) = 4n^2+n -4n^2+7n +3 =8n +3 =TN =8n +3 T1 = 8×1+3= 11 =a T2 =8×2 +3 = 19 d = T1-T2 =19 - 11= 8 =d AP = 11, 19, 27 ,.......

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    Roma Jebisow 1 year, 4 months ago

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