Question

if mth term of an A.P.is n and nth term is m, show that (m+n)th term vanishes

Answer 1

Answer:

m+n th term of  ap = 0

Step-by-step explanation:

if mth term of an A.P.is n and nth term is m, show that (m+n)th term vanishes

mth term of an A.P.  = a + (m-1)d = n  => a = n -(m-1)d

nth Term of an ap =  a + (n-1)d = m  => a = m - (n-1)d

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

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

=> n - m = d(m-n)

=> d = -1

Puttind d = -1

a = n + m - 1  

m+n th term of  ap = a + (m + n - 1)d

putting a = m+n-1  & d = -1

= n + m - 1 + (m + n  -1)(-1)

= n + m - 1 -m -n + 1

= 0

m+n th term of  ap = 0

Answer 2

According to the expression to find nth term of an AP

$\boxed{a_n = a + (n - 1)d}$
here a= first term

d = common difference

So ATQ

$\boxed{n = a + (m - 1)d} \: \: \: ......eq1$
$\boxed{m = a + (n - 1)d} \: \: \: \: eq2$
So (m+n)th term,find either the value of a or d,it is convenient to find the value of d,by subtracted both equations

Now subtract eq2- eq1

$m-n = a + (n - 1)d- a - (m - 1)d\\\\m-n= a-a + nd-d-md+d\\\\m - n =(n-m)d \\ \\ d= -\frac{m-n}{m-n} \\ \\ d=-1$

Now value of a can be find from eq1 or eq2

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

(m+n)th term

$\boxed{a_{m+n} = a + (m+n - 1)d} \\ \\=a+(m+n-1)(-1)\\\\=m+n-1-(m+n-1)\\\\=m+n-1-m-n+1\\\\=0$