Question

If 6 times the sixth term of an arithmetic progression is equal to 9 times the 9th term, find the 15th term.

Answer 1

Here we will first assume the first term the common difference of an AP to be any variable. Then we will form the equation using the condition given in the question. After simplifying the equation, we will get the first terms in terms of common differences. We will substitute the value of the first term in the expression for the 15th term to get required value.

Formula used:
${n^{th}}$term of an arithmetic progression is given by ${a_n} = a + \left( {n - 1} \right) \times d$, where $a$ is the first term of an AP, $d$ is the common difference and ${a_n}$ is the ${n^{th}}$term of an arithmetic progression.

Complete step-by-step answer:
Let the first term of an arithmetic progression be $a$ and let the common difference be $d$.
We will first find the 6th terms of an arithmetic progression using the formula ${a_n} = a + \left( {n - 1} \right) \times d$. Therefore,
${a_6} = a + \left( {6 - 1} \right) \times d$
Subtracting the terms inside the brackets, we get
$\Rightarrow {a_6} = a + 5d$ …………….. $\left( 1 \right)$
Now, we will find the 9th terms of an arithmetic progression using the formula ${a_n} = a + \left( {n - 1} \right) \times d$.
${a_9} = a + \left( {9 - 1} \right) \times d$
Subtracting the terms inside the brackets, we get
$\Rightarrow {a_9} = a + 8d$ …………….. $\left( 2 \right)$
As it is given that the 9 times the 9th term of an arithmetic progression is equal to 6 times the sixth term of an arithmetic progression
So we can mathematically write it as:-
$6{a_6} = 9{a_9}$
Now, we will substitute the value of the sixth term and ninth term of an arithmetic progression. Therefore, we get
$\Rightarrow 6\left( {a + 5d} \right) = 9\left( {a + 8d} \right)$
Now, multiplying the terms on both sides of the equation, we get
$\Rightarrow 6a + 6 \times 5d = 9a + 9 \times 8d$
Multiplying the terms, we get
$\Rightarrow 6a + 30d = 9a + 72d$
Rewriting the terms, we get
$\Rightarrow 30d - 72d = 9a - 6a$
On subtracting the like terms, we get
$\Rightarrow - 42d = 3a$
Now, we will divide both sides by 3. Therefore, we have
\Rightarrow \dfrac{{ - 42d}}{3} = \dfrac{{3a}}{3} \\\ \Rightarrow - 14d = a \\\
$\Rightarrow a = - 14d$ ……….. $\left( 3 \right)$
Now, we will find the 15th term of an arithmetic progression.
${a_{15}} = a + \left( {15 - 1} \right) \times d$
On further simplification, we get
$\Rightarrow {a_{15}} = a + 14d$
Now, substituting $a = - 14d$ obtained in equation $\left( 3 \right)$ in the above equation, we get
$\Rightarrow {a_{15}} = - 14d + 14d$
On subtracting the terms, we get
$\Rightarrow {a_{15}} = 0$
Hence, the value of the 15th term of an AP is equal to 0.

Note: Here we have obtained the 15th term of the given AP. We know that AP stands for Arithmetic Progression and it is defined as the sequence of the numbers which are in order and in which the difference of any two consecutive numbers is a constant value. A real-life example of AP is when we add a fixed amount to our money bank every week. Similarly, when we ride a taxi, we pay an amount for the initial kilometer and pay a fixed amount for all the further kilometers, this also turns out to be an AP.