If m times the mth term of and APis equal to the n times its nth term find the (m+n)th term of the AP
Answers
Answer:
Thge general nth term of an ap is a+(n-1)d.
From the given conditions,
m(a+(m-1)d)=n(a+(n-1)d) => am+m2d-md=an+n2d-nd
=>a(m-n) +(m+n) (m-n) d-(m-n) d=0
=>(m-n)(a+(m+n)-1)d=0
Rejecting the non-trivial case of m=n are different.
=>(a+(m+n-1) d)=0
The LHS of this equation denotes the (m+n) th term of the ap, which is zero.
Hope it helps....
Please mark me as brainliest.............
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