Question

determine the A.P.whose third term is 9 and and when fifth term is subtracted from 8th term we get 6??

Answer 1

Answer:5,3,1.....

Step-by-step explanation:

See the atachment

Answer 2

Given

$a_3=9$

$a_8-a_5=6$

To find

The A.P.

Solution

$a_n=a+(n-1)d$

$a_3=a+(2-1)d\\\Rightarrow 9=a+d$ ...(i)

$a+(8-1)d-[a+(5-1)d]=6\\\Rightarrow a+7d-[a+4d]=6\\\Rightarrow a+7d-a-4d=6\\\Rightarrow 3d=6\\\Rightarrow d=\dfrac{6}{3}=2$

Substituting in the first equation we get

$9=a+d\\\Rightarrow a=9-2\\\Rightarrow a=7$

The AP is $7,9,11,13,15,17,19,21\ .....$

We can check the calculations $a_8-a_5=21-15=6$ so it is verified.