Question

If (m + 1)th term of an A.P is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.

Answer 1

Answer:

Step-by-step explanation:

Given :  

(m + 1)th term = 2 (n + 1)th term  

Let 'a' be the first term of an A.P and 'd' is the common difference of an A P.

Now ,

(m + 1)th term = 2 (n + 1)th term  

By using the formula , nth term ,an = a + (n -1)d

a +(m + 1 - 1)d = 2{a + (n + 1 -1)d }

a + md = 2{a + nd}

a + md = 2a + 2nd

md - 2nd = 2a - a

(m - 2n)d = a …………...(1)

We have to prove that   (3m + 1)th term  =  2 (m + n + 1)th term

Now,

LHS = (3m + 1)th term  

= a + (3m + 1 -1)d

= a + 3md

= (m - 2n) d + 3md  

[From eq 1]

= md - 2nd + 3md

= md + 3md - 2nd

= 4md - 2nd

LHS = 2d(2m - n)

 

RHS = 2 (m + n + 1)th term

= 2 {a + (m + n + 1 - 1)d}

= 2 {a + (m + n)d}

= 2 {(m - 2n)d + (m + n) d}

[From eq 1]

= 2{d(m - 2n + m + n)}

= 2{d(2m - n)}

RHS = 2d(2m - n)  

LHS = RHS

Hence proved.

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Answer 2

SOLUTION

Given,

(m+1)th term=a+(m+1-1)d=a+m.d.......(1)

(n+1)th term=a+(n+1-1)d=a+n.d.........(2)

Now Given condition is

=) a+md=2(a+n.d).........(3)

Now,

(3m+1)th term=a+(3m+1-1)d=a+3m.d..(4)

(m+n+1)th term=a+(m+n+1-1)d=a+(m+n)d....(5)

Now from (4)

(3m+1)th term = a+md+2md.....(6)

from (3) we have

=) a+md= 2(a+nd)..... put in (6)

(3m+1)th term =2(a+nd)+ 2md

=) 2(a+nd+md)

=) 2{a+(m+n)d}

=) (3m+1)th term = 2(m+n+1)th [from(5)]

Hence proved

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