Question

If $m$ times the $m^{th}$ term of an A.P is equal to $n$ times its $n^{th}$ term, then find its $(m+n)^{th}$ term.

Answer 1

Answer:

( m + n )th term of the AP is 0.

Step-by-step explanation:

It is given that the m times of $m^{th}$ term is equal to the n times of $n^{th}$ term.

So,

⇒ m times of $m^{th}$ term = n times of $n^{th}$

We know,

xth term of the AP = a + ( x - 1 )d, wher a is the 1st term, x is the number of term and d is the common difference between them .

Now,

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

= >  m x [ a + dm - d ] = n x [ a + dn - d ]

= >  am + dm^2 - dm = an + dn^2 - dn

= >  am - an + dm^2 - dm^2 = dm - dn

= >  a( m - n ) + d( m^2 - n^2 ) = d( m - n )

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

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

= >  a + d( m + n ) = d

= >  a + dm + dn = d

= >  a + dm + dn - d = 0

= >  a + d( m + n - 1 ) = 0

= >  a + ( m + n - 1 ) d = 0    ...( i )

From the identity, given above,

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

From ( i ) , value of a + ( m + n - 1 )d is 0, so the value of ( m + n )th term of the AP is 0.

Answer 2

Step-by-step explanation:

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!