Question

If m times of the mth term is equal to n times of nth term then prove that {m+n}th = 0. ​

Answer 1

Answer:

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Answer 2

$\huge\mathfrak\green{AnsweR}$

According to question.

m × tm = n × tn

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

ma +m(m-1)d = na +n(n-1)d

m(m-1)d - n(n-1)d = na - ma

[m²-m - n² + n]d = a(n-m)

[m²-n² -m+n ]d = a(n-m)

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

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

-a = [m+n-1)d

Now,

${t}_{m+n} = a+(m+n-1)d$

${t}_{m+n} = a - a [(m+n-1)d = -a]$

${t}_{m+n} = 0$

Hence Proved!!!