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