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Hint- Here, we will be using the formula for finding the nth term of an arithmetic progression in order to write the expressions of 9th term, 29th term and 19th term of an arithmetic progression.

To prove- 29th term is double of 19th term of given AP i.e., ${a_{29}} = 2{a_{19}}$

Let us suppose the first term of an AP as ${a_1}$, common difference as d.

As we know that the nth term of any AP with first term as ${a_1}$ and common difference as d is given by ${a_n} = {a_1} + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$

Given, ${a_9} = 0$

Using equation (1), we have

${a_9} = {a_1} + \left( {9 - 1} \right)d \Rightarrow {a_1} + 8d = 0{\text{ }} \to {\text{(2)}}$

Let us simplify the equation which needs to be proved by putting n=29 in equation (1) for LHS and n=19 in equation (1) for RHS, we can write

$

{a_{29}} = 2{a_{19}} \Rightarrow {a_1} + \left( {29 - 1} \right)d = 2\left[ {{a_1} + \left( {19 - 1} \right)d} \right] \Rightarrow {a_1} + 28d = 2\left( {{a_1} + 18d} \right) \Rightarrow {a_1} + 28d = 2{a_1} + 36d \\

\Rightarrow {a_1} + 8d = 0{\text{ }} \to {\text{(3)}} \\

$

Clearly, the equation which needs to be proved is reduced into equation (3) so in order to prove the required equation ${a_{29}} = 2{a_{19}}$, equation (3) needs to be proved.

Since, equation (2) holds true and equation (3) needs to be proved whereas equations (2) and (3) are the same. Hence, the required equation is proved i.e., ${a_{29}} = 2{a_{19}}$(29th term of AP is double of 19th term of AP).

Note- In these types of problems, we will simplify the equation that needs to be proved with the help of general formulas for an arithmetic progression and then use the already given condition. In this particular problem the given condition is the same equation (obtained after simplification) which needs to be proved.

To prove- 29th term is double of 19th term of given AP i.e., ${a_{29}} = 2{a_{19}}$

Let us suppose the first term of an AP as ${a_1}$, common difference as d.

As we know that the nth term of any AP with first term as ${a_1}$ and common difference as d is given by ${a_n} = {a_1} + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$

Given, ${a_9} = 0$

Using equation (1), we have

${a_9} = {a_1} + \left( {9 - 1} \right)d \Rightarrow {a_1} + 8d = 0{\text{ }} \to {\text{(2)}}$

Let us simplify the equation which needs to be proved by putting n=29 in equation (1) for LHS and n=19 in equation (1) for RHS, we can write

$

{a_{29}} = 2{a_{19}} \Rightarrow {a_1} + \left( {29 - 1} \right)d = 2\left[ {{a_1} + \left( {19 - 1} \right)d} \right] \Rightarrow {a_1} + 28d = 2\left( {{a_1} + 18d} \right) \Rightarrow {a_1} + 28d = 2{a_1} + 36d \\

\Rightarrow {a_1} + 8d = 0{\text{ }} \to {\text{(3)}} \\

$

Clearly, the equation which needs to be proved is reduced into equation (3) so in order to prove the required equation ${a_{29}} = 2{a_{19}}$, equation (3) needs to be proved.

Since, equation (2) holds true and equation (3) needs to be proved whereas equations (2) and (3) are the same. Hence, the required equation is proved i.e., ${a_{29}} = 2{a_{19}}$(29th term of AP is double of 19th term of AP).

Note- In these types of problems, we will simplify the equation that needs to be proved with the help of general formulas for an arithmetic progression and then use the already given condition. In this particular problem the given condition is the same equation (obtained after simplification) which needs to be proved.

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