Question

If S_r, denotes the sum of the first r terms of an AP then, \(\dfrac{S_{3r} - S_{r - 1}}{S_{2r} - S_{2r-1} }\)is
1. 2r + 1
2. 2r + 3
3. 2r - 1
4. 4r + 1

Answer 1

Correct Answer - Option 1 : 2r + 1

Concept:

We know that, 

S_n =  (n/2) × [ 2A + (n - 1) d]

Where S_n is the sum of n terms of A.P

n is the number of the terms 

A is the first term, d is a common difference.

Calculation:

Let a is the first term and d is the common difference of A.P

Such that ,S_n = (n/2) × [2a + (n - 1) d]  

and S_n - S_n-1 = T_n

So, S_2r - S_{2r - 1} = T_2r

S_3r -  Sr _{- 1} = 3r/2 × [2a + (3r -1) d] - r-1/2 × [2a + (r - 2) d]

⇒ 2a[3r/2 - (r - 1)/2] +(d/2) × [(9r² - 3r) - (r²  - 3r + 2)]

⇒ 2a × 2r + 1/2 + d (4r² - 1)

⇒ (2r + 1) [2a + (2r - 1) d]

And = ( 2r + 1 ). T_2r

⇒ (S3r  - Sr -1) / ( S2r  - S2r  _{- 1  =}  ( 2_r + 1 ) ×( T_2r  / T_2r)

∴ ( 2r + 1 )