# Given that 2,a,b the terms of a G.P. and 2,17,a the terms of an A.P. calculate a, b.

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2,a ,b are in GP............(1)

2,17 , a are in AP..........(2)

Solution:

Since 2 , a and b are in GP,

a/2 = b/a common ratio of consecutive terms in GP.

a^2 = 2b...........(3).

From (2) , 2, 17 and a are in AP.

So the common difference between the consecutive terms are equal.So 17 -2 = a-17.

17+17-2 = a .

32 = a.

a = 32. Substitute a = 32 in the relation at (3):

a^2 = 2b , 32^2 = 2b

b = 32^2/2 = 512.

Therefore a = 32 , b = 512.

I

Here I am assuming the they are the consecutive terms of the series.

Using the fact that 2,a,b are the terms of a GP. a/2=b/a or a^2=2b.

Using the fact that 2,17,a are the terms of an AP, 17=(2+a)/2 or 2+a=34 or a=32.

As we know that a^2=32^2=2b, b=32^2/2= 32*32/2=32*16=512

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

If 2,a,b are in geometric progression, then we'll apply the theorem of geometric progressions: each term of a geometric progression, beginning with the second term, is the geometric mean between the 2 joined terms:

a^2=2*b

If 2,17,a are in arithmetic progression, then the middle term is the arithmetic mean between the 2 joined terms:

17 = (a+2)/2

a+2=17*2

**a=32**

But a^2=2*b => 32^2 = 2*b

We'll divide by 2:

32*16 = b

** b=512**