Find the sum of the following arithmetic progression: x - y x + y , 3 x - 2 y x + y , 5 x - 3 y x + y , … to n terms.

Given that, the sequence x - y x + y , 3 x - 2 y x + y , 5 x - 3 y x + y , . . . to n terms is in A.P. Here, first term a = x - y x + y and common difference d = 3 x - 2 y x + y - x - y x + y = 2 x - y x + y We know that, the sum of first n terms of the A.P is S n = n 2 [ 2 a + n - 1 d ] ∴ S n = n 2 2 x - y x + y + n - 1 2 x - y x + y ⇒ S n = n 2 x + y 2 x - y + n - 1 2 x - y ⇒ S n = n 2 x + y 2 x - 2 y + 2 nx - 2 x - yn + y ⇒ S n = n 2 x + y n 2 x - y - y Hence, the sum of first n terms of the A.P is n 2 x + y n 2 x - y - y .

Find the sum of the series: 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 + … to 3 n terms.

Given 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 + … to 3 n term This can be written as sum of three alternate terms such as 3 + 6 + 9 + … t o n t e r m + 5 + 9 + 13 + … t o n t e r m + 7 + 12 + 17 + … t o n t e r m So, clearly, the above series is in Arithmetic Progression i..e, first series, second series and third series So, for first series: first term = 3 and common difference = 6 - 3 = 3 And, for second series: first term = 5 and common difference = 9 - 5 = 4 Similarly, for third series: first term = 7 and common difference = 12 - 7 = 5 We know that, S n = n 2 2 a + ( n - 1 ) d = n 2 2 × 3 + n - 1 × 3 + n 2 2 × 5 + n - 1 × 4 + n 2 2 × 7 + n - 1 × 5 = n 2 3 + 3 n + 6 + 4 n + 9 + 5 n = n 2 12 n + 18 = n 6 n + 9 = 3 n 2 n + 3 Hence, the sum of the given series 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 + … to 3 n is 3 n ( 2 n + 3 ) .

How many terms of the A . P . - 6 , - 11 2 , - 5 , … are needed to given the sum - 25 ?

Given A . P . is - 6 , - 11 2 , - 5 Here, a = - 6 and d = 1 2 Let Total number of terms are n ∴ S n = n 2 2 a + n - 1 d ⇒ n 2 2 × - 6 + n - 1 1 2 = - 25 ⇒ - 25 × 2 = n - 12 + n 2 - 1 2 ⇒ - 50 = n n 2 - 12 - 1 2 ⇒ - 50 = n n 2 - 12 ( 2 ) + 1 2 ⇒ - 50 = n n 2 - 24 + 1 2 ⇒ - 50 = n n 2 - 25 2 ⇒ n n - 25 = - 100 ⇒ n 2 - 25 n + 100 = 0 ⇒ n 2 - 20 n - 5 n + 100 = 0 ⇒ n ( n - 20 ) - 5 ( n - 20 ) = 0 ⇒ n - 20 n - 5 = 0 ⇒ n = 20 or n = 5 Hence, n = 5 , 20 both values gives the required sum.

E M B I B E

MATHEMATICS for CLASS XI>Chapter 19 - Arithmetic Progressions>EXERCISE 19.4

R D Sharma Solutions for Chapter: Arithmetic Progressions, Exercise 4: EXERCISE 19.4

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Arithmetic Progressions, Exercise 4: EXERCISE 19.4

Attempt the practice questions on Chapter 19: Arithmetic Progressions, Exercise 4: EXERCISE 19.4 with hints and solutions to strengthen your understanding. MATHEMATICS for CLASS XI solutions are prepared by Experienced Embibe Experts.