Q. 3 D5.0( 1 Vote )

# For a given A.P. withT10 = 41, S10 = 320, find Tn, Sn.

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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