Question

The seventh term of an arithmetic progression is four times its second term and twelfth term is $2$ more than three times of its fourth term. Find the progression.

Answer 1

In this question we will first try to make equations based on the relationships we have been given with the terms in the progression and then we will solve them to get the final progression.
We will make use of the $n^{th}$ term formula which is given by
The ${n^{th}}$ term of the arithmetic progression $a,a + d,a + 2d.....a + nd$ is
${t_{_n}} = a + (n - 1)d$
Here, ${t_n}$ is the ${n^{th}}$ term in the progression, $a$ is the first term in the progression, and $n$ is the total number of terms in the progression
Also $d$ is the common difference in the progression.

Complete step-by-step answer:
In this question stated as the given that the seventh term in the arithmetic progression is four times the second term in the same progression,
Here we have to write in mathematically by using the formula, ${t_7} = 4 \times {t_2}$
Now the seventh term in the progression is $a + 6d$ and the second term is $a + d$ therefore, it can be written as: $a + 6d = 4(a + d)$
On opening the bracket on the right-hand side, we get:
$\Rightarrow a + 6d = 4a + 4d$
On taking similar terms across the $=$ sign we get:
$\Rightarrow 6d - 4d = 4a - a$
On subtracting we get
$\Rightarrow 2d = 3a$,
On taking as equated form we get
$\Rightarrow 3a - 2d = 0 \to (1)$
Also, in the question stated as the twelfth term is $2$ more than three times of its fourth term, mathematically it can be written as:
${t_{12}} = 3 \times {t_4} + 2$
Now the twelfth term in the progression is $a + 11d$ and the fourth term is $a + 3d$, therefore it can be written as:
$a + 11d = 3(a + 3d) + 2$
On multiply the bracket, we get:
$\Rightarrow$ $a + 11d = 3a + 9d + 2$
On taking similar terms across the $=$ sign we get:
$\Rightarrow$ $a - 3a = 9d - 11d + 2$
On simplifying we get:
$\Rightarrow$ $- 2a = - 2d + 2$
Therefore,
$\Rightarrow$ $2a - 2d = - 2 \to (2)$
On subtracting $(1) - (2)$ we get:
$\Rightarrow$ $a = 2$
Now substituting the value of $a = 2$ in $(1)$ we get:
$\Rightarrow$ $3(2) - 2d = 0$
On multiplication we get,
$\Rightarrow$ $2d = 6$
Let us divide the terms and we get
$\Rightarrow$ $d = 3$.
Therefore, the required arithmetic progression is:
$2, 2 + 3,2 + 6,2 + 9.........$
On adding the terms and we get
$2, 5, 8, 11........$
Hence we get the required answer.

Note: To check whether the value is correct or not, we will substitute the values found in the conditions given in the question.
The seventh term of an arithmetic progression is four times its second term
${t_7} = 4 \times {t_2}$
$a + 6d = 4(a + d)$
On substituting the values, we get:
$2 + 6 \times 3 = 4(2 + 3)$
On simplification we get
$2 + 18 = 4(5)$
On adding and multiply the terms we get,
$20 = 20$
Since left-hand side $=$ right-hand side, the values are correct.
Now, twelfth term is $2$ more than three times of its fourth term
${t_{12}} = 3 \times {t_4} + 2$
$a + 11d = 3(a + 3d) + 2$
On substituting the values, we get:
$2 + 11 \times 3 = 3(2 + 3 \times 3) + 2$
Let us multiply the terms and we get,
$2 + 33 = 3(2 + 9) + 2$
Let us adding we get,
$35 = 3(11) + 2$
On multiply the terms and adding we get
$35 = 35$
Since left-hand side $=$ right-hand side, the values are correct.