Question

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

Answer 1

Hint: We will find the (m+1)th term of AP then we will find the (n+1)th term of the AP then we will build an equation for this relation and we will solve for (3m+1)th term and then it will be twice that of (m+n+1)th term.
Formula used:
The nth term of an AP is given by the formula $ {a_n} = a + (n - 1)d $ where, a is the first term and d is the common difference.

Complete step-by-step answer:
Given (m+1)th term of an AP is twice of (n+1)th term, Let a be first term & d be common difference $ {a_m} + 1 = 2{a_{n + 1}} $
 $ a + (m + 1 - 1)d = 2\left[ {a + \left( {n + 1 - 1} \right)d} \right] $ ​
 $ \Rightarrow a + md = 2a + 2nd $
 $ \Rightarrow md - 2nd = 2a - a $
 $ \Rightarrow d(m - 2n) = a $
 $ \Rightarrow d = \dfrac{a}{{(m - 2n)}} $ --(1)
Now (3m+1)th term of AP is
 $ {a_{3m + 1}} = a + \left( {3m + 1 - 1} \right)d = a + 3md $
Putting value of d from (1) we get
 $ {a_{3m + 1}} = a + 3m\left( {\dfrac{a}{{(m - 2n)}}} \right) $
 $ \Rightarrow {a_{3m + 1}} = \dfrac{{ma - 2na + 3ma}}{{m - 2n}} $
 $ \Rightarrow {a_{3m + 1}} = \dfrac{{4ma - 2na}}{{m - 2n}} $ --(2)
Now (m+n+1)th term of AP is
 $ {a_{m + n + 1}} = a + \left( {m + n + 1 - 1} \right)d $
 $ \Rightarrow {a_{m + n + 1}} = a + \left( {m + n} \right)d $
 $ \Rightarrow {a_{m + n + 1}} = a + \left( {m + n} \right) \times \dfrac{a}{{(m - 2n)}} $
 $ \Rightarrow {a_{m + n + 1}} = \dfrac{{am - 2an + am + an}}{{m - 2n}} $
 $ \Rightarrow {a_{m + n + 1}} = \dfrac{{2am - an}}{{m - 2n}} $
 $ \Rightarrow {a_{m + n + 1}} = \dfrac{1}{2} \times \dfrac{{4am - 2an}}{{m - 2n}} = \dfrac{1}{2}{a_{3m + 1}} $
 $ \Rightarrow {a_{3m + 1}} = 2{a_{m + n + 1}} $
So, the correct answer is “ $ {a_{3m + 1}} = 2{a_{m + n + 1}} $ ”.

Note: While solving this we should take care that we are applying the correct formula of AP. The first term of an AP is a constant term and the difference too is a constant, the only thing that varies throughout an AP are the terms.