Question

The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by \[\frac{l^2 - a^2}{k - (l + a)}\] , then k =

 

 

Answer 1

In the given problem, we are given the first, last term, sum and the common difference of an A.P.

We need to find the value of k

Here,

First term = a

Last term = l

Sum of all the terms = S

Common difference (d) =  `(l^2 - a^2)/(k - ( l +a ))`

Now, as we know,

l =a + ( n -1) d                       .....................(1)

Further, substituting (1) in the given equation, we get

` d = ([a + (n-1) d]^2 - a^2)/( k - {[a + (n -1)d] + a})`

`d = (a^2 +[(n-1)d]^2 + 2a (n-1)d - a^2)/(k - {[a +(n-1)d] + a})`

` d = ([(n - 1)d]^2 + 2a (n-1)d)/(k - {[a + (n -1)d ] +a})`

Now, taking d in common, we get,

                         ` d = ([(n-1)d]^2 + 2a(n -1)d)/(k - {[a +(n-1)d]+ a})`

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

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

Taking (n-1) as common, we get,

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

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

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

Further, multiplying and dividing the right hand side by 2, we get,

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

Now, as we know,  `S = n/2 [(n - 1) d + 2a]`

Thus,

k = 2S