Question

if m times of mth term of an ap is equal to n times of nth term then show that (m+n)th term of the ap is zeros

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}mth 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

here is your answer OK ☺☺☺☺☺



Let a be the first terms and d be the common difference of given A.P.

Then, am = a + (m – 1) d and an = a + (n – 1) d

According to the given condition, m times the mth term of an a.p is equal to n times the nth term of an a.p.

⇒ m × am = n × an

⇒ m × [a + (m – 1) d] = n [a + (n – 1) d]

⇒ am + dm (m – 1) = an + dn (n – 1)

⇒ am – an = dn (n – 1) – dm (m – 1)

⇒ a(m – n) = d(n 2 – n – m 2 + m)



Now, (m + n)th term of an a.p = am + n = a + (m + n – 1) d

⇒ (1 – m – n)d + (m + n – 1)d [Using (1)]

⇒ d – md – nd + md + nd – d

⇒ 0

Hence, the (m + n)th terms of given AP is zero.

Hope you help ☺☺☺