Let the first term of the arithmetic sequence be a and the common difference be d.

(a + a + 12d)*(13/2) = 130

=> 2a + 12d = 20

=> a + 6d = 10

=> a = 10 - 6d

a4, a10 and a7 are consecutive terms of a...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Let the first term of the arithmetic sequence be a and the common difference be d.

(a + a + 12d)*(13/2) = 130

=> 2a + 12d = 20

=> a + 6d = 10

=> a = 10 - 6d

a4, a10 and a7 are consecutive terms of a geometric series:

=> (a + 9d) / (a + 3d) = (a + 6d) / (a + 9d)

=> (10 - 6d + 9d) / (10 - 6d + 3d) = (10 - 6d + 6d) / (10 - 6d + 9d)

=> (10 + 3d) / (10 - 3d) = 10 / (10 + 3d)

=> 100 + 9d^2 + 60d = 100 - 30d

=> 9d^2 + 90d = 0

=> d( d + 10) = 0

=> d = 0 and d = -10

So the first term of the series = 10 - 6*d can be 10 or 70

**The required value of the first terms of the AP can be 10 or 70.**