Question

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

Answer 1

Please refer to attachment.

Answer 2

Given :

• m times the $\sf{m^{th}}$ of an A.P. is equal to n times its $\sf{n^{th}}$ term.

To prove :

• The $\sf{(m+n)^{th}}$ term A.P. is 0.

Solution :

★ m the term A.P. →

$\sf{T_m=a+(m-1)d}$

★ n th term A.P. →

$\sf{T_n=a+(n-1)d}$

According to the question :-

$\sf{m[a+(m-1)d]=n[a+(n-1)d]}$

$\to\sf{ma+(m^2-m)d=na+(n^2-n)d}$

$\to\sf{ma-na+(m^2-m)d-(n^2-n)d=0}$

$\to\sf{a(m-n)+d(m^2-m-n^2+n)=0}$

$\to\sf{a(m-n)+d[(m+n)(m-n)-1(m+n)=0}$

$\to\sf{(m-n)[a+d(m+n-1)]=0}$

$\to\sf{a+d(m+n-1)=0}$[m≠n]

$\to\sf{T_{m+n}=0\:\{Proved\}}$

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Some formulas :-

★ Number of terms in an AP ,

n= [ (l-a) / d ] + 1

• [where n = number of terms,a= the first term,l = last term,d= common difference.]

★ Sum of first n term in an AP,

Sn = n/2 [2a + (n − 1) d]

• [where,a = the first term, d= common difference ]