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Question: If \(S_{r}\) denotes the sum of the first *r* terms of an A.P., then \(\frac{S_{3r} - S_{r - 1}}{S_{...

If SrS_{r} denotes the sum of the first r terms of an A.P., then S3r−Sr−1S2r−S2r−1\frac{S_{3r} - S_{r - 1}}{S_{2r} - S_{2r - 1}} is equal to

A

2r – 1

B

2r + 1

C

4r + 1

D

2r + 3

Answer

2r + 1

Explanation

Solution

S3r−Sr−1S2r−S2r−1=3r2{2a+(3r−1)d}−(r−1)2{2a+(r−1−1)d}T2r=(2r+1)a+d2{3r(3r−1)−(r−1)(r−2)}a+(2r−1)d=(2r+1)a+d2{8r2−2}a+(2r−1)d=(2r+1)a+d(4r2−1)a+(2r−1)d=2r+1\frac{S_{3r} - S_{r - 1}}{S_{2r} - S_{2r - 1}} = \frac{\frac{3r}{2}\{ 2a + (3r - 1)d\} - \frac{(r - 1)}{2}\{ 2a + (r - 1 - 1)d\}}{T_{2r}} = \frac{(2r + 1)a + \frac{d}{2}\{ 3r(3r - 1) - (r - 1)(r - 2)\}}{a + (2r - 1)d} = \frac{(2r + 1)a + \frac{d}{2}\{ 8r^{2} - 2\}}{a + (2r - 1)d} = \frac{(2r + 1)a + d(4r^{2} - 1)}{a + (2r - 1)d} = 2r + 1