Question

If m times the mth term of an AP is equal to n times its nth term.show that the (m+n)th term of AP is zero

Answer 1

Answer:

T(m+n)=0

Step-by-step explanation:

°let first term of an A.P is a and common difference is d.

NOW,,

• m.Tm=n.Tn
• m.[a+(m-1).d]=n.[a+(n-1).d]
• am+(m-1).d.m=an+(n-1).d.n
• am+(m-1).d.m -- an -(n-1).d.n= 0
• a(m-n)+d.{(m-1).m- (n-1).n}=0
• a.(m-n)+d .[ m2 - m - n2 + n ]=0

We will use formula.

a2-b2 =(a+b).(a-b)

SO,,

• a(m-n)+ d. [ (m+n)(m-n) - (m-n) ] = 0
• a(m-n) + d. [ ( m-n) {(m+n-1)} ] =0

Taking m- n common from above equation ,,

• (m-n). [ a + (m+n-1).d ]=0
• [a+(m+n-1).d]=0
• T(m+n) = 0

proved.

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