Question

15. If m times the mth term of an A.P. is equal to n times its nth term, then show that the
(m+n)th term of the A.P. is zero.

Answer 1

Hi ,

Let the first term and the common difference

of the A.P be a and d respectively.

mth term = a + ( m - 1 )d

nth term = a + ( n - 1 )d

according to the problem given ,

m times mth term = n times of nth term

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

ma + ( m² - m )d = an + ( n² - n )d

ma + m²d - md = an + n² d - nd

ma - an + m² d - n² d - 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

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

a + ( m + n )d - d = 0

a + ( m + n - 1 )d = 0

( m + n ) th of term = 0

I hope this helps you.

: )

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!