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Two series X (x1, x2, x3, x4, ….xn) and Y (y1, y2, y3, y4, ….., yn) are in A.P., such that xn – yn = n − 2. It is also known that x3= b5. Find the value of x99 – y197

(a) 47 

(b) 48 

(c) 49 

(d) 50

Sarthaks Test For Class 7-12

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Best answer

Correct option  (b) 48

Explanation: 

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

Since xn – yn = 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.

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