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 =

विकल्प

• 1

• 2

• 3

• none of these.

Answer 1

In the given problem, we are given  d = S_n - kS_n-1 + S_n-2

We need to find the value of k

So here,

First term = a

Common difference = d

Sum of n terms = S_n

Now, as we know,

`S_n = n/2 [2a + (n - 1) d ] `             .............(1) 

Also, for n-1 terms,

`S_(n-1) = (n-1)/2 [ 2a + [(n-1) - 1]d]`

          ` = (n-1)/2 [2a + ( n -2) d ] `                  .............(2) 

Further, for n-2 terms,

`S_(n - 2) = (n - 2)/2 [ 2a + [(n - 2 ] d]`

         `= (n-2)/2 [ 2a + ( n - 3) d ] `          ...............(3) 

Now, we are given,

                d = S_n  - kS_{n -1} + S_n-2 

d  +  KS_n-1 = S_n + S_n-2

               ` K = (S_n + S_(n-2) - d ) /S_(n-1)`

Using (1), (2) and (3) in the given equation, we get

`k = (n/2[2a + (n-1) d ] + (n-2)/2 [2a + ( n - 3)d]-d)/((n-1)/2 [ 2a + (n - 2 ) d ])`

Taking `1/2` common, we get,

`K = (n[2a + (n-1)d] + (n-2) [2a + ( n- 3)d]- 2d)/((n-1)[2a + (n-2) d])`

    ` =(2an + n^2d - nd + 2an + n^2d - 3nd - 4a - 2nd + 6d - 2d)/(2an + n^2d - 2nd - 2a - nd + 2d)`

 ` = (2n^2d + 4an - 6nd - 4a + 4d)/(n^2d + 2an - 3nd + -2a + 2d)`

Taking 2 common from the numerator, we get,

`k = (2(n^2 d + 2an - 3nd  + -2a + 2d))/(n^2d + 2an - 3nd + -2a +2d)`

= 2

Therefore, k = 2