Q. 3 D5.0( 1 Vote )

For a given A.P. with

T10 = 41, S10 = 320, find Tn, Sn.

Answer :

Formula used.


Sn = [a + an]


Tn = an = a + (n–1)d


Sn = [2a + (n–1)d]


S10 = 320


T10 = a10 = 41


S10 = [a + 41]


320 = [a + 41]


= a + 41


64 = a + 41


a = 64–41


a = 23


a10 = a + (n–1)d


41 = 23 + (10–1)d


41–23 = 9d


9d = 18


d = = 2


Tn = an = a + (n–1)d


Tn = 23 + (n–1)2


23–2 + 2n


21 + 2n


Sn = [2a + (n–1)d]


Sn = [2×23 + (n–1)2]


= [46 + 2n–2]


= [44 + 2n]


= n[22 + n]


= 22n + n2


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