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

Answer:

The value of k is 2.

Among the given options option (b) 2 is a correct answer.

Step-by-step explanation:

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

Given :  

d = Sn – k Sn–1 + Sn – 2

Let number of terms , n = 3

So, AP is : a, (a + d), (a + 2d)

& d = S3 – k S3 –1 + S3–2

d = S3 – k S2 + S1 ..............(1)

 

By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]

On putting n = 2 in eq 1,

S1 = 1/2 [2a + (1 – 1) d]

S1 = ½ [2a + 0]

S1 = ½ × 2a

S1 = a

On putting n = 2 in eq 1,

S2 = 2/2 x [2a + (2–1) d]  

S2 = 1 [2a + d]

S2 = (2a + d)

On putting n = 3 in eq 1,

S3 = 3/2 x (2a + (3 –1)d)

S3 = 3/2 x (2a + 2d)

S3 = 3/2 x 2(a + d)

S3 = 3(a + d)

S3 = 3a + 3d

On putting the values of S1, S2 & S3 in eq 1,  

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(2a + d)/(2a + d)

k = 2

Hence, the value of k is 2.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Let n = 3

AP is : a , a + d, a + 2d

d = S3 - kS3-1 + S3-2  

  = S3 - kS2 + S1  

Sum of n terms of an AP is given as:  

Sn = (n/2)*{2a + (n-1)d}

Now S1 = a

S2 = (2/2)*(2a + (2-1)d)      

S2 = (2a + d)          

S3 = (3/2)*(2a + (3-1)d)  

S3 = (3/2)*(2a + 2d)

S3 = 3(a + d)

S3 = 3a + 3d

Put value of S1 , S2 and S3

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