Question

13. If m times the term of an A.P. is eqaul to n times n term then show that the
(m + n) term of the A.P. is zero.​

Answer 1

In AP, nth term = a + (n - 1)d.  According to the question:

⇒ m times the term of an A.P. is equal to n times n term

⇒ m[a + (m - 1)d] = n[a + (n - 1)d]

⇒ ma + md(m - 1) = na + nd(n - 1)

⇒ ma - na = nd(n - 1) - md(m - 1)

⇒ a(m - n) = n²d - nd - m²d + md

⇒ a(m - n) = d[n² - m² - n + m]

⇒ a(m - n) = d[(n-m)(n+m)-(n-m)]

⇒ a(m - n) = d(n - m)(m + n - 1)

⇒ a = - d(m + m - 1)

⇒ a + d(m + n - 1) = 0

 Notice that (m + n)th term = a + (m + n - 1)d, which we have proved to be 0.

⇒ (m + n)th = a + (m + n - 1) = 0

  Proved

Answer 2

Solution :-

We know that

$\sf a_n = a(n +1)-d$

Now,

$\sf m\bigg(a + (m - 1) d\bigg) = n\bigg(a + (n - 1)d\bigg)$

$\sf ma + md \times (n-1) = na + nd \times (n - 1)$

$\sf ma + md (n-1) = na+nd(n-1)$

$\sf ma-na= nd(n-1)-md(n-1)$

Taking a as common

$\sf a(m - n) =nd(n-1)-md(n-1)$

$\sf a(m-n) = d(n^2 - m^2 - m - n)$

$\sf a(m + n) = d(n-m)(m+n-1)$

Cancellation of n

$\sf a = -d(m+m-1)$

$\sf a = -d(m +n-1)$

$\sf a + d = m +n-1$

$\sf a + d(m+n-1) = 0$