Question

If m times the mth term of an AP is equal to n times its nth term ,find the (m + n) th term of the AP

Answer 1

m(a + (m-1) d) = n ( a + (n-1) d).
ma+md(m-1) = na+ nd(n-1).
ma - na = n^2d - nd - m^2d + md.
a(m-n) = -d (m^2 - n^2 ) + d ( m-n).
a(m-n)= -d(m-n)(m+n) + d (m-n).
Using identity a^2 - b^2 = (a+b)(a-b).
a(m-n) = d(m-n)[-(m+n) + 1].
a = -d(m+n) + d
a + d (m+n) - d = 0 .
a +d(m+n-1) = 0.
Therefore value of m+ n th term is 0.
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Answer 2

Answer:

We know : an = a +(n-1)d a (m+n) = a + (m+n-1)d (just put m+n in place of n ) --------------(1)

Let the first term and common difference of the A.P. be ‘a’ and ‘d’ respectively. Then,

m th term = a + (m – 1) d and n th term = a + (n-1)d

By the given condition,

↠ m x am = n x an m [a + (m – 1) d] = n [a + (n – 1) d]

↠ ma + m (m – 1) d = na + n (n – 1) d

↠ ma + (m2 -m)d - na - (n2 -n)d = 0 ( taking the Left Hand Side to the other side)

↠ ma -na + (m2 - m)d -( n2-n)d = 0 (re-ordering the terms)

↠ a (m-n) + d (m2-n2-m+n) = 0 (taking 'a ' and 'd ' common)

↠ a (m-n) + d {(m+n)(m-n)-(m-n)} = 0 (a2-b2 identity) Now divide both sides by (m-n)

↠ a (1) + d {(m+n)(1)-(1)} = 0

↠ a + d (m+n-1) = 0 ---------------(2)

• From equation number 1 and 2 :

⇒ a (m+n) = a + (m+n-1)d

And we have shown : a + d (m+n-1) = 0

∴ a (m+n) = 0