Q. 3 D5.0( 1 Vote )

# For a given A.P. with

T_{10} = 41, S10 = 320, find T_{n}, S_{n}.

Answer :

Formula used.

S_{n} = [a + a_{n}]

T_{n} = a_{n} = a + (n–1)d

S_{n} = [2a + (n–1)d]

⇒ S_{10} = 320

⇒ T_{10} = a_{10} = 41

S_{10} = [a + 41]

320 = [a + 41]

= a + 41

64 = a + 41

a = 64–41

⇒ a = 23

a_{10} = a + (n–1)d

41 = 23 + (10–1)d

41–23 = 9d

9d = 18

⇒ d = = 2

T_{n} = a_{n} = a + (n–1)d

T_{n} = 23 + (n–1)2

23–2 + 2n

21 + 2n

S_{n} = [2a + (n–1)d]

S_{n} = [2×23 + (n–1)2]

= [46 + 2n–2]

= [44 + 2n]

= n[22 + n]

= 22n + n^{2}

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