Sum of n Terms of Series :
Here we are going to see some practice questions on finding sum of n terms of the series.
Question 1 :
Find the general term and sum to n terms of the sequence
1, 4/3 , 7/9 , 10/27, . . . .
Solution :
1, 4/3 , 7/9 , 10/27, . . . .
a = 1, d = 3, r = 1/3
General term :
tn = ((a + (n − 1)d)rn−1)
tn = ((1 + (n − 1)3)(1/3)n−1)
= (1 + 3n - 3)(1/3)n−1
tn = (3n - 2)/3n −1
Sn = 3n - (3n - 2)/2⋅3n-1 + (3n-1 - 1)/(8 ⋅ 3n-3)
Question 2 :
Find the value of n, if the sum to n terms of the series
√3 + √75 + √243 + · · · is 435√3.
Solution :
√3 + √75 + √243 + · · ·+ n terms
√3 + 5√3 + 9√3 + · · ·+ n terms
The given series is arithmetic series.
Sn = 435√3
(n/2)[2a+ (n-1)d] = 435√3
a = √3, d = 4√3
(n/2)[2√3+ (n-1)(4√3)] = 435√3
(n/2)[2√3+ 4√3n - 4√3)] = 435√3
(n/2)[4√3n - 2√3)] = 435√3
n[2√3n - √3] = 435√3
Divide by √3 on both sides
n[2n - 1] = 435
2n2 - n - 435 = 0
(n - 15) (2n + 29) = 0
n = 15, n = -29/2
Hence the value of n is 15.
Question 3 :
Show that the sum of (m + n)th and (m − n)th term of an AP. is equal to twice the mth term
Solution :
tm+n = a + (m + n - 1)d ------(1)
tm-n = a + (m - n - 1)d ------(2)
(1) + (2)
tm+n + tm-n = a + (m + n - 1)d + a + (m - n - 1)d
= a + md + nd - d + a + md - nd - d
= 2a + 2md - 2d
= 2 [a + (m-1) d]
= 2 tm
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