Question

if m times the mth term of an AP is equal to n times the nth term then show that (m+n)th term of an AP is 0

Answer 1

We know that,

m[ a+(m-1)d ] = n[ a+(n-1)d ]

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

am+m2d-md-an-n2d+nd =0

a(m-n) + [m2-n2-(m-n)]d=0 a(m-n)+{(m+n)(m-n)-(m-n)}d=0  

(m-n){a+(m+n-1)d} =0

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

Hence proved.

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Answer 2

Given m * tm = n * tn

m*(a + (m – 1)*d) = n*[a + (n – 1)*d]

m*a + m*(m – 1)*d = n*a + n*(n – 1)*d

=>m*a + m(m – 1)*d – n*a – n*(n – 1)*d = 0

a*(m – n) + (m2 – m – n2 + n)*d = 0

a*(m – n) + (m2 – n2 – m +  n)*d = 0

a*(m – n) + [(m – n)*(m + n) – (m –  n)]*d = 0

(m – n)[a+ {(m + n) – 1}]d = 0

[a+ (m + n) – 1]*d = 0

so that (m + n) th term = 0