Question

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

Answer 1

Answer:

we have a and d

now we can determine the ap

Answer 2

Answer:-

Let the first term of the AP = a and common difference = d, third term =16

$\small\sf{a3=16}$ = $\small\sf{a+2d=16}$ .........(1)

7th term exceeds the 5th term by 12.

$\large\sf{Therefore,}$

$\small\sf{a7=a5+12}$

$\small\sf{a+6d=a+4d+12}$

$\small\sf{2d=12}$

$\small\sf{d=6}$

Putting the value of d in equation 1,we get

$\small\sf{a+2d(6)=16}$

$\small\sf{a=4}$

Hence,the AP = a, a+d, a+2d... = 4,10,16.