2 terms is possible, how about 3?

Algebra Level 3

Does there exist an arithmetic progression, with a positive first term a a and a positive common difference d d such that the first three terms of the arithmetic progression are also the first three terms of a geometric progression?

No Yes, for all values of a a Yes, but only for certain values of a a

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2 solutions

Mehul Gajwani
Jun 23, 2016

I think an easier solution is to treat the first three terms of an AP as if they are a GP:

a + d a = a + 2 d a + d ( a + d ) 2 = a ( a + 2 d ) \displaystyle \begin{aligned} \frac{a+d}{a} &= \frac{a+2d}{a+d} \\ \therefore (a+d)^2 &= a(a+2d) \end{aligned}

Simplifying this gives d 2 = 0 d = 0 d^2 = 0 \Rightarrow d = 0 , which is not positive.

Therefore there is no solution.

Simple and nice! Well done!

Hung Woei Neoh - 4 years, 11 months ago
Hung Woei Neoh
May 28, 2016

Now, let us assume that the first 3 3 terms of an AP are also the first 3 3 terms of a GP.

The AP: a , a + d , a + 2 d , a, a+d, a+2d,\ldots

The GP: a , a r , a r 2 , a,ar,ar^2,\ldots

Since the first 3 3 terms are the same,

a + d = a r a+d = ar \implies Eq.(1)

a + 2 d = a r 2 a+2d = ar^2\implies Eq.(2)

Eq.(1) × 2 \times 2 - Eq.(2):

2 ( a + d ) ( a + 2 d ) = 2 a r a r 2 a = 2 a r a r 2 2(a+d) - (a+2d) = 2ar-ar^2\\ a=2ar-ar^2

Since a > 0 a >0 , we can divide the equation by a a :

1 = 2 r r 2 r 2 2 r + 1 = 0 ( r 1 ) 2 = 0 r 1 = 0 r = 1 1=2r-r^2\\ r^2-2r+1=0\\ (r-1)^2=0\\ r-1=0\\ r=1

For the first 3 3 terms of an AP and GP to be the same, r r must be 1 1 . However,

a + d = a ( 1 ) d = a a = 0 a+d=a(1)\\ d=a-a=0

If r = 1 r=1 , d = 0 d=0

This means that for an AP and GP to share the same first 3 3 terms, the common difference must be 0 0 .

This in turn implies that for an AP with a positive common difference, the first 3 3 terms of the sequence cannot be the first 3 3 terms of a GP.

The answer is No \boxed{\text{No}}

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