Question

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

Answer 1

Hey there !!

Given :-

→ ( m·a$\tiny m$ ) = ( n·a$\tiny n$ ) .

To prove :-

→ ( m + n )th  [ a$\tiny m + n$ ] = 0.

Solution :-

Let a be the first term and d be the common difference of the given AP.

Then,

a$\tiny m$ = a + ( m - 1 )d.

And,

a$\tiny n$ = a + ( n - 1 )d.

Now,

→ ( m·a$\tiny m$ ) = ( n·a$\tiny n$ ) .

⇒ 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 ) + d· { ( m² - n² ) - ( m - n ) } = 0.

⇒ ( m - n ) · { a + ( m + n - 1 )}d = 0.

⇒ ( m - n ) · a$\tiny m + n$ = 0.

⇒ a$\tiny m + n$ = 0.

Hence, it is proved.

THANKS

#BeBrainly.

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!