Question

Let S_n denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = S_n – kS_n–1 + S_n–2, then k = 

A. 1 

B. 2 

C. 3 

D. none of these.

Answer 1

Correct answer is B. 2

Let a be the first term, n be the number of terms and d be the common difference of AP. 

Given 

d = S_n – kS_n–1 + S_n–2. 

Now let n = 3 

So, AP is : a, a + d, a + 2d 

And 

d = S₃ – k S_3–1 + S_3–2 

d = S₃ – k S₂ + S₁ ..............(1)

Sum of n terms of an AP is given as: 

S_n = (\(\frac{n}{2}\)) x {2a + (n–1) d} 

Now S₁ = a 

S₂ = (\(\frac{2}{2}\)) x (2a + (2–1) d) (n =2) 

S₂ = (2a + d) 

S₃ = (\(\frac{3}{2}\)) x (2a + (3–1) d) (n =3) 

S₃ = \(\frac{3}{2}\) x (2a + 2d) 

S₃ = 3(a + d) 

S₃ = 3a + 3d 

Putting values of S₁, S₂ and S₃ in equation 1, we get

d = 3a + 3d – k (2a + d) + a 

d = 4a + 3d – k (2a + d) 

k (2a + d) = 4a + 3d – d 

k (2a + d) = 4a + 2d 

k (2a + d) = 2(2a + d) 

k = 2