Question

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

Answer 1

Let first term of AP = a

Let common difference of AP = d

It is given that its term is equal to 16. It means $a_3$=16, where $a_3$ is the $3^{rd}$ term of AP.

Using formula $a_n$=a+ (n−1) d, to find $n^{th}$ term of arithmetic progression, we can say that

$16=a+(3-1)\left(d\right)$

$\Rightarrow 16=a+2d$

It is also given that $7^{th}$ term exceeds $5^{th}$ term by12. Again using formula $a_n$ = a + (n−1) d, which is used to find $n^{th}$ term of arithmetic progression, we can say that

$a_7=a+(7-1)d=a+6d$ and, $a_5=a+(5-1)d=a+4d$ (1)

According to the given condition, we can say that

_{$a_7$} = $a_5$+$12$

Putting (1) in the above equation, we get

$\Rightarrow a+6d=a+4d+12$

$\Rightarrow 2d=12$

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

Putting value of d in equation 16=a+2d, we get

$16=a+2\left(6\right)$

$\Rightarrow a=4$

Therefore first term = a = 4

And, common difference = d = 6

Therefore, AP is 4, 10, 16, 22....