Question

The 5th and 9th term of an arithmetic progression are 7 and 13 respectively. what is the 15th term

Answer 1

Hey,sup!

As per the question,

Let the terms in AP be a,a+d,a+2d,a+3d....

So given , a+4d=7 and a+8d=13.

Subtracting 5th term from 9th term we get,

=> a+8d-a-4d=13-7.
=> 4d=6.
=> d=6/4= 1.5

Substituting d in 5th term
=> a+4d=7
=> a+4×1.5=7.
=> a+6=7.
=> a=7-6=1.

So 15th term will be, a+14d=1+14×1.5=1+21=22.

Hope it helps.

Answer 2

Answer:

Step-by-step explanation:

A5=a+4d

A5=7

7=a+4d___(1)

A9=a+8d

A9=13

13=a+8d___(2)

Eq1-2

a+4d=7

-a+8d=13

__________

-4d=-6

d=-6/-4

d=3/2

Put in eq 1

7=a+4d

7=a+4(3/2)

7=a+6

a=1