(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) to 11 terms: find the sum
Solution:
Given, the series is \(\frac{a-b}{a+b}+\frac{3a-2b}{a+b}+\frac{5a-3b}{a+b}+......\)
We have to find the sum of the arithemtic series upto 11 terms.
From the given series,
First term, a = (a-b)/(a+b)
Common difference, d = (3a-2b)/(a+b) - (a-b)/(a+b)
= (3a-2b-a+b)/(a+b)
d = (2a-b)/(a+b)
The sum of the first n terms of an AP is given by
Sₙ = n/2[2a + (n-1)d]
So, S₁₁ = (11/2)[2(a-b)/(a+b) + (11-1)(2a-b)/(a+b)]
= (11/2)[(2a-2b)/(a+b) + (10)(2a-b)/(a+b)]
= (11/2)[(2a-2b)/(a+b) + (20a-10b)/(a+b)]
= (11/2)[2a - 2b + 20a - 10b]/(a+b)
= (11/2)/(a+b)[22a - 12b]
Taking out common term,
= (11/2)/(a+b)[2(11a - 6b)]
= (11/a+b)[11a - 6b]
= 11(11a - 6b)/(a+b)
Therefore, the sum of the series upto 11 terms is 11(11a - 6b)/(a+b)
✦ Try This: Find the sum: \(\frac{a+b}{a-b}+\frac{3a+2b}{a-b}+\frac{5a+3b}{a-b}+......\)to 11 terms
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5
NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 21 (iii)
(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) to 11 terms: find the sum
Summary:
The sum of the series \(\frac{a-b}{a+b}+\frac{3a-2b}{a+b}+\frac{5a-3b}{a+b}+......\)to 11 terms is 11(11a - 6b)/(a+b)
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