Question

in an Ap ,if mth term is n and nth term is m,show that its rth term is ( m+ n - r)

Answer 1

mth term: a+(m-1)d=n --------1
nth term:  a+(n-1)d=m ---------- 2
now from 1 and 2 find a and d 
i.e a+md-d=n
a=n-md+d
ok now substitute this a value in the second equation to get the value of  d
so n-md+d+dn-d=m
d(n-m)=-(n-m)
i.e d=-1
now value of a=n+m-1
now since rth term is a+(r-1)d
= n+m-1+(r-1)-1
=m+n-1-r+1
= m+n-r
hence rth term is (m+n-r)
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Answer 2

Answer :

Let, the first term of the AP is a and the common difference is d.

Then,

mth term, n = a + (m - 1)d ...(i)

and

nth term, m = a + (n - 1)d ...(ii)

On subtraction, we get

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

⇒ n - m = (m - n)d

⇒ d = - 1

Putting d = - 1 in (i), we get

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

⇒ n = a - m + 1

⇒ a = m + n - 1

Thus, the rth term

= a + (r - 1)d

= m + n - 1 + (r - 1) (- 1)

= m + n - 1 - r + 1

= m + n - r

Hence, proved.

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