Question

If a,b,c are in AP and |a|,|b|,|c|<1, and $x = 1 + a + {a^2} + ... + \infty$ , $y = 1 + b + {b^2} + ... + \infty$ and $z = 1 + c + {c^2} + ... + \infty$, then x,y,z, shall be in
1). AP
2). GP
3). HP
4). None of these

Answer 1

Here in the given question, we have the expression which says about the values of variables given, and then need to find the relation between the variables, here we know that the variables already are in AP, and thus solve accordingly.
Formulae Used: Sum of infinite series which is in GP and magnitude of common difference is less then, is given by:

$$\Rightarrow x = 1 + a + {a^2} + ... + \infty \\\ \Rightarrow sum = \dfrac{1}{{1 - a}} \\\$$

Complete step-by-step solution:
Here in the given question, three series given, which is equal to the variables in the question, to solve further we need to first solve the series and then move forward and we know that the variables are in AP, so on solving we get:
Here we first know that all the three series given are in GP and we know the sum of infinite series which is in GP, and the magnitude of the common difference is less then one:

$$\Rightarrow x = 1 + a + {a^2} + ... + \infty = \dfrac{1}{{1 - a}} \\\ \Rightarrow y = 1 + b + {b^2} + ... + \infty = \dfrac{1}{{1 - b}} \\\ \Rightarrow z = 1 + c + {c^2} + ... + \infty = \dfrac{1}{{1 - c}} \\\$$

Here we know:
$\Rightarrow a,b,c\,are\,in\,AP$
So,
$\Rightarrow 1 - a,1 - b,1 - c\,are\,in\,AP$
Therefore;
$\Rightarrow \dfrac{1}{{1 - a}},\dfrac{1}{{1 - b}},\dfrac{1}{{1 - c}}\,will\,be\,in\,HP$
(Reciprocal of common difference of the series in AP, will be the common ratio for HP series)
Hence the required answer for the given question is:
$\Rightarrow x,y,z\,are\,in\,HP$

Note: Here we know that the ratio of HP is the reciprocal of the common difference, of the series which is in AP, and accordingly here we solved the question. Here we also see that if three numbers are in AP, then if every term is reduced by one will also be in AP.