Find the sum of n terms of the series 2 a 2 - 1 a , 4 a - 3 a , 6 a 2 - 5 a , ⋯ ⋯

The difference of the consecutive terms is constant, so the given sequence is an arithmetic progression. 1 st term A = 2 a 2 - 1 a . 2 nd term = 4 a - 3 a = 4 a 2 - 3 a . ∴ Common difference, d = 4 a 2 - 3 a - 2 a 2 - 1 a = 2 a a 2 - 1 . Using, sum of n terms, S n = n 2 [ 2 A + ( n - 1 ) d ] we get, S n = n 2 2 2 a 2 - 1 a + ( n - 1 ) a 2 - 1 a · 2 = n 2 a 2 - 1 a + ( n - 1 ) a 2 - 1 a = n a 2 a 2 - 1 + ( n - 1 ) a 2 - 1 = a n ( n + 1 ) - n 2 a

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Higher Algebra>Chapter 4 - Arithmetical Progression>Examples. IV.a.

H S Hall and S R Knight Solutions for Chapter: Arithmetical Progression, Exercise 1: Examples. IV.a.

Author:H S Hall & S R Knight

H S Hall Mathematics Solutions for Exercise - H S Hall and S R Knight Solutions for Chapter: Arithmetical Progression, Exercise 1: Examples. IV.a.

Attempt the free practice questions on Chapter 4: Arithmetical Progression, Exercise 1: Examples. IV.a. with hints and solutions to strengthen your understanding. Higher Algebra solutions are prepared by Experienced Embibe Experts.