Question

If m times mth term of an ap ia n times nth term of the ap then find the( m+n)th term

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.