Question

If in an A.P. S_n = n²p and S_m = m²p, where S_r denotes the sum of r terms of the A.P., then S_p is equal to 

विकल्प

• \[\frac{1}{2} p^3\]

   

• m n p

•  p³

• (m + n) p²

Answer 1

In the given problem, we are given an A.P whose `S_n = n^2 p` and  `S_m =  m^2 p`

We need to find S_P

Now, as we know,

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

Where, first term = a

Common difference = d

Number of terms = n

So,

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

`n^2 p  = n/2 [ 2a + (n-1)d]`

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

Similarly,

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

`m^2 p = m/2 [2a + (m + 1)d]`

     `p = 1/(2m)[2a + md -d] `             ...............(2)

Equating (1) and (2), we get,

\[\frac{1}{2n}\left( 2a + nd - d \right) = \frac{1}{2m}\left( 2a + md - d \right)\]

\[\Rightarrow m\left( 2a + nd - d \right) = n\left( 2a + md - d \right)\]

\[\Rightarrow 2am + mnd - md = 2an + mnd - nd\]

Solving further, we get,

2am - 2an = - nd + md 

2a ( m - n) = d (m - n)

             2a = d                      ..............(3) 

Further, substituting (3) in (1), we get,

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

`n^2 p   = n/2 [d + nd - d ]`

      `p = 1/(2n)[nd]`

     `p = d/2`                  ..............(4) 

Now,

`S_p = p/2 [2a + ( p - 1) d ]`

`S_p = p/2 [ d +pd - d] `               ( Using 3)

`S_p = p/2 [ p(2 p)] `                         ( Using 4 )

`S_p = p^3`

Thus, `S_p = p^3`