lf m times the mth term of an Ap is equal to n times its n th term ,find its (m+ n) th term.
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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