Question

if (m+1)th term of an AP is twice the (n+1)th term proe that (3n+1)th term is twice the ( m+n+1)th ​

Answer 1

Answer:

THEREFORE , (3m + 1 ) the term is twice the ( m + n + 1) the term.

Step-by-step explanation:

$(m + 1)th \: \: term = a + (md) = 2 \times ((n + 1)th \: \: term) \\ \\ a + md = 2(a + nd) \\ \\ a + md = 2a + 2nd \\ \\ md = a + 2nd \\ \\ a = (m - 2n)d \\ \\ \\ \\ (m + n + 1)th \: \: term = a + (m + n)d \\ \\ = (m - 2n)d + md + nd \\ \\ = 2md - nd \\ \\ \\ \\ and \: \: \: \: (3m + 1) \: th \: \: term = a + 3md \\ \\ = (m - 2n)d + 3md \\ \\ =4 md - 2nd \\ \\ = 2(2md - nd) \\ \\ = 2 \: times \: (m + n + 1) \: th \: term.$