if (m+1)th term of an AP is twice the n+1th term, prove that (3m+1)th term is twice the (m+n+1)th term.

please answer ASAP, exam tmrw.

write all the given conditions as

(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)

form (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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