Question

If the sum of m terms of an A.P. is equal to the sum of either the next n terms or the next p terms, then prove that `(m + n) (1/m - 1/p) = (m + p) (1/m - 1/n)`

Answer 1

Let the A.P. be a, a + d, a + 2d, ...

We are given

a₁ + a₂ + ... + a_m = a_m+1 + a_m+2 + ... + a_m+n  .....(1)

Adding a₁ + a₂ + ... + am on both sides of (1), we get

2[a₁ + a₂ + ... + a_m] = a₁ + a₂ + ... + a_m + a_m+1 + ... + a_m+n

2 S_m = S_m+n

Therefore, `2 m/2{2a + (m - 1)d} = (m + n)/2 {2a + (m + n - 1)d}`

Putting 2a + (m – 1) d = x in the above equation, we get

mx = `(m + n)/2` (x + nd)

(2m – m – n)x = (m + n)nd

⇒ (m – n)x = (m + n) nd   ....(2)

Similarly, if a₁ + a₂ + ... + a_m = a_{m + 1} + a_{m + 2} + ... + a_{m + p}

Adding a₁ + a₂ + ... + a_m on both sides

we get, 2(a₁ + a₂ + ... + a_m) = a₁ + a₂ + ... + a_{m + 1} + ... + a_{m + p}

or 2 S_m = S_{m + p}

⇒ `2 m/2 {2a + (m - 1)d} = (m + p)/2 {2a + (m + p - 1)d}` which gives

i.e., (m – p) x = (m + p)pd   ....(3)

Dividing (2) by (3), we get

`((m - n)x)/((m - px)) = ((m + n)nd)/((m + p)pd)`

⇒ (m – n)(m + p)p = (m – p)(m + n)n

Dividing both sides by mnp, we get

`(m + p)(1/n - 1/m) = (m + n) (1/p - 1/m)`

= `(m + n) (1/m - 1/p) = (m + p) (1/m - 1/n)`