 QUESTION

# If ${\left( {m + 1} \right)^{th}}$ term of an A.P. is twice the ${\left( {n + 1} \right)^{th}}$ term, prove that ${\left( {3m + 1} \right)^{th}}$ term is twice the ${\left( {m + n + 1} \right)^{th}}$ term.

Hint: For solving such questions of A.P. we don’t need to go by basics. Rather we will use some common formulas of Arithmetic Progression taking the help of given problem statements.

Let the first term of the given Arithmetic Progression be $a$ and the common difference be $d$.
To prove ${\left( {3m + 1} \right)^{th}}$term is twice the ${\left( {m + n + 1} \right)^{th}}$ term.
We are given that ${\left( {m + 1} \right)^{th}}$ term is twice the ${\left( {n + 1} \right)^{th}}$ term.
${\left( {m + 1} \right)^{th}}$ term of the given A.P. is
${T_{m + 1}} = a + \left( {m + 1 - 1} \right)d\left[ {\because {T_n} = a + (n - 1)d} \right] \\ = a + md \\$
${\left( {n + 1} \right)^{th}}$ term of the given A.P. is
${T_{n + 1}} = a + \left( {n + 1 - 1} \right)d\left[ {\because {T_n} = a + (n - 1)d} \right] \\ = a + nd \\$
Given in the problem: that ${\left( {m + 1} \right)^{th}}$ term is twice the ${\left( {n + 1} \right)^{th}}$ term
$\Rightarrow {T_{m + 1}} = 2{T_{n + 1}}$
Substituting the value of ${\left( {m + 1} \right)^{th}}$ term and the ${\left( {n + 1} \right)^{th}}$ term in the above equation and solving
$\Rightarrow a + md = 2\left( {a + nd} \right) \\ \Rightarrow a = d\left( {m - 2n} \right) \\ \Rightarrow a = md - 2nd \\ \Rightarrow 2nd = md - a \\ \Rightarrow nd = \dfrac{{md - a}}{2}................(1) \\$
Now, as we know that ${\left( {3m + 1} \right)^{th}}$ term is
${T_{3m + 1}} = a + \left( {3m + 1 - 1} \right)d\left[ {\because {T_n} = a + (n - 1)d} \right] \\ = a + 3md........(2) \\$
Also, we have ${\left( {m + n + 1} \right)^{th}}$ term:
${T_{m + n + 1}} = a + \left( {m + n + 1 - 1} \right)d\left[ {\because {T_n} = a + (n - 1)d} \right] \\ = a + \left( {m + n} \right)d \\ = a + md + nd.......(3) \\$
Now substituting the value of equation (1) in equation (3)
${T_{m + n + 1}} = a + md + nd \\ = a + md + \dfrac{{md - a}}{2} \\$
Multiplying both sides by 2
$\Rightarrow 2{T_{m + n + 1}} = 2a + 2md + md - a \\ = a + 3md \\$
Which is same as equation (2)
So $\Rightarrow 2{T_{m + n + 1}} = {T_{3m + 1}}$
Hence if ${\left( {m + 1} \right)^{th}}$ term of an A.P. is twice the ${\left( {n + 1} \right)^{th}}$ term, then the ${\left( {3m + 1} \right)^{th}}$term is twice the ${\left( {m + n + 1} \right)^{th}}$ term.

Note: In order to solve any problem related to arithmetic progression, just consider some first term and common differences to start solving the problem. The formula for ${n^{th}}$ term of an A.P. and sum of n terms is very important and must be remembered. In order to solve these types of problems just find some common terms between the given part and the part to be proved and make the substitution.