Question

Determine an A.P. whose 3rd term is 16 and the difference of 7th term and 5th term is 12.

Answer 1

Here, we will find the common difference by usinng the formula of $n$th term of the arithmetic progression A.P., that is, ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term and $d$ is the common difference. Apply this formula, and then substitute the value of $a$,$d$ and $n$ in the obtained equation to find the required A.P.

Complete step by step answer:

We are given that the 3rd term is 16 and the difference of 7th term and 5th term is 12.
We know that the arithmetic progression is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.
$\Rightarrow {a_7} - {a_5} = 12$
Using the formula of $n$th term of the arithmetic progression A.P., that is, ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term and $d$ is the common difference, in the above equation, we get

$$\Rightarrow \left( {a + 6d} \right) - \left( {a + 4d} \right) = 12 \\\ \Rightarrow a + 6d - a - 4d = 12 \\\ \Rightarrow 2d = 12 \\\$$

Dividing the above equation by 2 on both sides, we get

$$\Rightarrow \dfrac{{2d}}{2} = \dfrac{{12}}{2} \\\ \Rightarrow d = 6 \\\$$

Using the above value of $d$ in the 3rd term ${a_3} = a + 2d$, we get

$$\Rightarrow {a_3} = a + 2\left( 6 \right) \\\ \Rightarrow {a_3} = a + 12 \\\$$

Taking ${a_3} = 16$ in the above equation, we get
$\Rightarrow 16 = a + 12$
Subtracting the above equation by 12 on both sides, we get

$$\Rightarrow 16 - 12 = a + 12 - 12 \\\ \Rightarrow 4 = a \\\ \Rightarrow a = 4 \\\$$

Hence the first term of A.P. is 4.
Substituting these values of $a$ and $d$ in the above formula for the second term of the arithmetic progression, we get

$$\Rightarrow {a_2} = 4 + 6 \\\ \Rightarrow {a_2} = 10 \\\$$

Hence, the A.P. is 4, 10, 16, …

Note: In solving these types of questions, you should be familiar with the formula of sum of the arithmetic progression and their sums. Some students use the formula of sum, $S = \dfrac{n}{2}\left( {a + l} \right)$, where $l$ is the last term, but have the to find the value of ${a_n}$ , so it will be wrong. We can also find the value of $n$th term by find the value of ${S_n} - {S_{n - 1}}$, where ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $a$ is the first term and $d$ is the common difference. But this is a longer method, which takes time, so we will use the above method. One should know the ${a_n}$ is the $n$th term in the arithmetic progression.