Question

if n times the nth term an AP is equal to m times the mth term ,then (m+n)th term is zero



kindly explain this instead of solving ​

Answer 1

Step-by-step explanation :

Given,

n times the nth term an AP is equal to m times the mth term

To prove,

(m+n)th term is zero

Formula,

nth term of AP is given by,

➣ aₙ = a + (n - 1)d

where

a is the first term

d is the common difference

Solution,

To prove that (m+n)th term is 0,

• Find the mth and nth term by substituting in the above formula.

• Then, substitute the values of mth and nth terms in the given relation : n times the nth term an AP is equal to m times the mth term

• By simplifying, we'll get (m + n)th term

--------------------------------------

So,

➯ nth term of AP is,

➙   aₙ = a + (n - 1)d

➯ mth term of AP is,

➙   aₘ = a + (m - 1)d

According to the given relation,

 $\boxed{\sf \textbf{n times the nth term = m times the mth term}}$

      n × aₙ = m × aₘ

Substituting the values of aₙ and aₘ ;

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

n [ a + nd - d ] = m [ a + md - d ]

na + n²d - nd = ma + m²d - md

ma + m²d - md - na - n²d + nd = 0

ma - na + m²d - n²d - md + nd = 0

 a(m - n) + d (m² - n²) - d(m - n) = 0

 a(m - n) + d [ (m - n) (m + n) ] - d(m - n) = 0       [ ∵ a² - b² = (a - b)(a + b) ]

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

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

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

   a + (m + n - 1)d = 0/(m - n)

   a + (m + n - 1)d = 0     [ ∵ 0 ÷ any number = 0 ]

    aₘ₊ₙ = 0

Therefore, (m + n)th term is 0

Hence proved!

Answer 2

Answer:

Step-by-step explanation:

To prove that (m+n)th term is 0,

Find the mth and nth term by substituting in the above formula.

Then, substitute the values of mth and nth terms in the given relation : n times the nth term an AP is equal to m times the mth term

By simplifying, we'll get (m + n)th term

--------------------------------------

So,

➯ nth term of AP is,

➙   aₙ = a + (n - 1)d

➯ mth term of AP is,

➙   aₘ = a + (m - 1)d

According to the given relation,

     n × aₙ = m × aₘ

Substituting the values of aₙ and aₘ ;

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

n [ a + nd - d ] = m [ a + md - d ]

na + n²d - nd = ma + m²d - md

ma + m²d - md - na - n²d + nd = 0

ma - na + m²d - n²d - md + nd = 0

a(m - n) + d (m² - n²) - d(m - n) = 0

a(m - n) + d [ (m - n) (m + n) ] - d(m - n) = 0       [ ∵ a² - b² = (a - b)(a + b) ]

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

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

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

  a + (m + n - 1)d = 0/(m - n)

  a + (m + n - 1)d = 0     [ ∵ 0 ÷ any number = 0 ]

   aₘ₊ₙ = 0

Therefore, (m + n)th term is 0

hope it helps..... :D