Question

The numbers \[\sqrt 2 + 1\] , \[1\] , \[\sqrt 2 - 1\] will be in
A. A.P
B. G.P
C. H.P
D. None of these

Answer 1

Hint: Progressions (or Sequences and Series) are numbers arranged in a particular order such that they form a predictable order. By predictable order, we mean that given some numbers, we can find the next numbers in the series.

Complete step by step answer:
Arithmetic Progression (AP): A sequence of numbers is called an arithmetic progression if the difference between any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by adding a fixed number to the previous number in the series. This fixed number is called the common difference. If \[a\] is the first term and \[d\] is the common difference then the nth term of an AP is \[a + (n - 1)d\] .

Geometric Progression (GP): A sequence of numbers is called a geometric progression if the ratio of any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by multiplying a fixed number to the previous number in the series. This fixed number is called the common ratio.If \[a\] is the first term and \[r\] is the common ratio then the nth term of the GP is \[a{r^{n - 1}}\] .

Harmonic Progression(HP): A sequence of numbers is called a harmonic progression if the reciprocal of the terms are in AP. In simple terms \[a,b,c,d,e,f\] are in HP if \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d},\dfrac{1}{e},\dfrac{1}{f}\] are in AP.

Now the given numbers \[\sqrt 2 + 1\] , \[1\] , \[\sqrt 2 - 1\] do not have any common difference so they cannot be in an AP. Also since their reciprocals are not in AP therefore the given numbers are not in HP. Hence we will now check if they are in GP or not .
We know that if \[a,b,c\] are in a GP then \[\dfrac{b}{a} = \dfrac{c}{b}\] or \[{b^2} = ac\]
Let us check this condition for given numbers.
Here \[a = \sqrt 2 + 1\] , \[b = 1\] and \[c = \sqrt 2 - 1\]
Therefore we have \[{b^2} = 1\] and \[ac = \left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right) = 1\]
Therefore it satisfies the equation \[{b^2} = ac\]. Therefore the given terms are in GP.

Hence option B is the correct answer.

Note: A sequence of numbers is called an arithmetic progression if the difference between any two consecutive terms is always the same. A sequence of numbers is called a geometric progression if the ratio of any two consecutive terms is always the same. A sequence of numbers is called a harmonic progression if the reciprocal of the terms are in AP.