In an AP of Sn = 4n -n^2 find its 25th term
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Given, Sn = n (4n+1)
= 4n^2+ n
W.K.T,
Tn= Sn-S(n-1)
= 4n^ 2+n-4(n-1)^2-(n-1)
= 4(n^2-n^2+2n-1)+(n-n+1)
= 8n-4+1
= 8b-3
Hence,
Tn = 8n-3
T1 = 8(1)-3=5
T2=8(2)-3=13
So AP is 5,13,21 and so on...
Hope this helps you friend
Thanks ✌️ ✌️
= 4n^2+ n
W.K.T,
Tn= Sn-S(n-1)
= 4n^ 2+n-4(n-1)^2-(n-1)
= 4(n^2-n^2+2n-1)+(n-n+1)
= 8n-4+1
= 8b-3
Hence,
Tn = 8n-3
T1 = 8(1)-3=5
T2=8(2)-3=13
So AP is 5,13,21 and so on...
Hope this helps you friend
Thanks ✌️ ✌️
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