Question

Two series X (x_1, x_2, x_3, x_4, ….x_n) and Y (y_1, y_2, y_3, y_4, ….., y_n) are in A.P., such that x_n – y_n = n − 2. It is also known that x₃= b₅. Find the value of x₉₉ – y₁₉₇. 

(a) 47 

(b) 48 

(c) 49 

(d) 50

Answer 1

Correct option  (b) 48

Explanation: 

Let the common difference of the series X be d₁ and that of Y be d₂

Since x_n – y_n = n − 2, x₁ − y₁ = −1 or y₁ = x₁ + 1 

x₃ = y₅

x₁ + 2d₁ = y₁ + 4d₂

x₁ + 2d₁= x₁ + 1 + 4d₂

2d₁ − 4d₂ = 1 

x₉₉ – y₁₉₇ = x₁+ 98d₁ − y₁- 196d₂ = −1 + 49(2d₁− 4d₂) = −1 + 49 = 48.