Question

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Answer 1

Here we need to determine the AP. For that, we will first find the third term of an AP using the formula and then we will equate it with the given value of the third term. From there, we will get the first equation including first term and common difference of an AP and will first find the seventh term and 5th term of an AP using the formula and form the second equation including first term and common difference of an AP from the given information. On solving these two equations, we will get the value of first term and common difference and thus the required AP.

Complete step by step solution:
Let’s assume the AP be $a$, $a+d$, $a+2d$, …….
We know the formula for finding the terms of an AP is given by
$\Rightarrow {{T}_{n}}=a+\left( n-1 \right)d$
Now, we will find the third term of an AP using this formula.
$\Rightarrow {{T}_{3}}=a+\left( 3-1 \right)d$
On further simplification, we get
$\Rightarrow {{T}_{3}}=a+2d$
It is given that the value of third term of this AP is 16.
Now, we will substitute this value here.
$\Rightarrow a+2d=16$ ……… $\left( 1 \right)$
Now, we will find the 7th term of an AP using this formula.
$\Rightarrow {{T}_{7}}=a+\left( 7-1 \right)d$
On further simplification, we get
$\Rightarrow {{T}_{7}}=a+6d$ …….. $\left( 2 \right)$
Now, we will find the 5th term of an AP using this formula.
$\Rightarrow {{T}_{5}}=a+\left( 5-1 \right)d$
On further simplification, we get
$\Rightarrow {{T}_{5}}=a+4d$ ………. $\left( 3 \right)$
It is given that the 7th term exceeds the 5th term by 12.
Mathematically,
$\Rightarrow {{T}_{7}}={{T}_{5}}+12$
On substituting the value of ${{T}_{7}}$ and ${{T}_{5}}$ obtained in equation 2 and 3, we get
$\Rightarrow a+6d=a+4d+12$
Adding and subtracting the like terms, we get
$\Rightarrow 2d=12$
Dividing both sides by 2, we get
b$\Rightarrow d=6$
Now, we will substitute the value of $d$ in equation 1, we get
$\Rightarrow a+2\times 6=16$
On further simplification, we get
$\Rightarrow a+12=16$
On subtracting 12 from both sides, we get
$\begin{aligned} & \Rightarrow a+12-12=16-12 \\\ & \Rightarrow a=4 \\\ \end{aligned}$
We can find AP by adding $d$ continuously.
Therefore, required AP is 4, $4+6$, $4+2\times 6$,….
On further simplification, we get

Hence, the required AP is 4, 10, 16, 22, …….

Note:
We have obtained the terms of an arithmetic progression. An arithmetic progression is defined as a sequence in which the difference between the term and the preceding term is constant or in other words, we can say that an arithmetic progression is a sequence such that every element after the first is obtained by adding a constant term to the preceding element.