Primes from arithmetic and geometric progressions

Question

The five primes, 131, 157, 211, 349, 739, are neither in arithmetic or geometric progression, but are instead the sum of the corresponding terms of two progressions of five terms each, one arithmetic and one geometric.

What are those two progressions?

Answer 1

The series are:

The arithmetic series: $124, 136, 148, 160, 172$

and

The geometric series: $7, 21, 63, 189, 567$

Method:

I got them by putting the formulae for AP and GP and solving for them

Answer 2

SamRoy's answer is correct.

Here is a demonstration of how to compute the answer in practice

If the first term of the arithmetic sequence is $a$ and the common difference is $d$ then the five terms are $$a, a+d,a+2d, a+3d, a+4d$$ Similarly, if the first term of the geometric sequence is $b$ and the common ratio is $r$ then the terms in the geometric sequence are $$ b, br, br^2, br^3, br^4$$ So we have $$ a+b = 131 \Rightarrow a = 131-b$$ $$ a+d+br = 157 \Rightarrow (131-b)+d+br = 157 \Rightarrow d = 26+(1-r)b$$ $$ a+2d+br^2 = 211 \Rightarrow (131-b) + 2(26+(1-r)b) + br^2 = 211$$ $$ \Rightarrow 183+b-2br +br^2 = 211 \Rightarrow b = \frac{28}{(r-1)^2} \Rightarrow d = 26 - \frac{28}{r-1} $$ $$ \Rightarrow a = 131 - \frac{28}{(r-1)^2}$$ Finally, we have $$ a+3d + br^3 = 349 \Rightarrow \left(131 - \frac{28}{(r-1)^2}\right) + 3\left(26 - \frac{28}{r-1}\right) + \frac{28r^3}{(r-1)^2} = 349$$ $$ \Rightarrow \frac{28r^3 -84r+56}{(r-1)^2} = 140 \Rightarrow 28\left(\frac{(r-1)^2(r+2)}{(r-1)^2}\right) = 140$$ and because the expressions for $a$, $b$ and $d$ allow us to conclude that $r \neq 1$, we can divide above and below by $(r-1)^2$ to get $$ 28(r+2) = 140 \Rightarrow r = 3$$ From the above expressions, we then immediately get $$a = 124, b = 7, d = 12$$ and we just need to do the final check that $$ a + 4d + br^4 = 739 $$