Question

For a given A.P., T10=41 and S10=320. Find Tn and Sn for this A.P.

Answer 1

a10 = 41
41 = a + ( 10 - 1 ) d
41 = a + 9d eq.1

s10 = 320
10/2 ( 2a + ( 10-1 )(d) ) = 320
5 ( 2a + 9d ) = 320
2a + 9d = 64
a + 41 = 64. [from eq. 1 ]
a = 23
Putting value in eq.1 we get
41 = 23 + 9d
18 = 9d
d = 2
an = a + ( n-1 ) (d)
an = 23 + ( n-1 ) (2)
an = 23 + 2n -2
an = 21 + 2n

sn = n/2 [a + an ]
sn = n/2 [ 23 + 21 + 2n ]
sn = n/2 [ 44 + 2n ]
sn = n [ 22 + n ]
sn = 22 n + n^2

Answer 2

Step-by-step explanation:

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