# What are a,b, if 2,a,b are the terms of a g.p. and 2,17,a are the terms of an a.p.?

*print*Print*list*Cite

### 3 Answers

We are given that 2, a, b are the terms of a geometric progression and 2, 17 , a are the terms of an arithmetic progression.

Now for an A.P. , if we consider three terms, the sum of the first and third terms is twice the second term.

=> 2 + a = 2*17

=> 2 + a = 34

=> a = 32

Now the terms of the G.P. are 2, 32, b

=> (32 / 2) = (b / 32)

=> 16 = b / 32

=> b = 32* 16

=> b = 512

**Therefore a = 32 and b = 512.**

2, a, b are in GP and 2,17, a are in AP.

Consider the AP : 2,17, a. So the common difference d is 2nd term - 1st term = d = 17-2 = 15.

Therefore d = 15.

So the 3rd term a = 2nd term +d = 17+15 = 32.

So now Put a = 32 in the GP is 2, a, b. So the given GP is 2, 3, b.

=> the common ratio r of the GP is = 2nd term / first term = 32/2 = 16.

So the 3rd term of the GP is b = 2nd term * r = 32* 16 = 512.

Therefore a = 32 and b = 512.

If 2,a,b are the terms of a geometric series, then we can apply the geometric mean theorem.

a^2 = 2b

If 2,17,a are the terms of an arithmetical series, then we can apply the arithmetical mean theorem.

17 = (2+a)/2

34 = 2 + a

a = 34 - 2

a = 32

a^2 = 2b <=> 32^2 = 2b

1024 = 2b

b = 512

**The real numbers a and b are: a = 32 and b = 512.**