Question

The mth
and nth
terms of an A.P. are n and m
respectively . prove that (m+n)th term will be zero​

Answer 1

Step-by-step explanation:

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

nth term =a+(n-1)d=m.......... (2)

subtracting equation(1)from (2) we get

(m-n)d=n-m

d=-1..................... (3)

now dividing equation (1)by(2)we get

[a+(m-1)d]÷[a+(n-1)d]=n÷m........ (4)

put the value of d from equation (3)to (4) we get

[a+1-m]m=[a+1-n]n

am+m-m^2=an+n-n^2

a (m-n)+(m-n)-(m-n)(m+n)

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

since m-n could not be equal to zero

so, a=m+n-1........ (5)

(m+n)th term =a+(m+n-1)d

from equation (3)and (5) we get

(m+n)th term=a+a×-1=a-a=0 (proved ).

Answer 2

Answer:

Let the first term of AP = a

common difference = d

We have to show that (m+n)th term is zero or a + (m+n-1)d = 0

mth term = a + (m-1)d

nth term = a + (n-1) d

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

⇒ am + m²d -md = an + n²d - nd

⇒ am - an + m²d - n²d -md + nd = 0

⇒ a(m-n) + (m²-n²)d - (m-n)d = 0

⇒ a(m-n) + {(m-n)(m+n)}d -(m-n)d = 0

⇒ a(m-n) + {(m-n)(m+n) - (m-n)} d = 0

⇒ a(m-n)  + (m-n)(m+n -1) d  = 0

⇒ (m-n){a + (m+n-1)d} = 0 

⇒ a + (m+n -1)d = 0/(m-n)

⇒ a + (m+n -1)d = 0

Proved!