Question

What does the series \[1 + {3^{ - \dfrac{1}{2}}} + 3 + \dfrac{1}{{3\sqrt 3 }} + \ldots \] represents?
A. AP
B. GP
C. HP
D. None of the above series

Answer 1

Hint: If the reciprocal of the terms is in AP, a series of numbers is referred to as a harmonic progression. Simply said, if \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d},\dfrac{1}{e}\] are in AP, then \[a,{\rm{ }}b,{\rm{ }}c,{\rm{ }}d,{\rm{ }}e\] are in HP so in our case, we have to solve the series with all the progression formulas to check whether the given series is A.P, G.P or H.P to get the desired answer.

Formula Used: To determine whether it is A.P use
\[a + (n - 1)d\]
To determine whether it is G.P use
\[a{r^{n - 1}}\]
To determine whether it is H.P use
\[b = \dfrac{{2ac}}{{a + c}}\]

Complete step by step solution: We are provided in the question that the series is
\[1 + {3^{ - \dfrac{1}{2}}} + 3 + \dfrac{1}{{3\sqrt 3 }} + \ldots \]
And are asked to determine what the progression is.
A mathematical progression is said to be geometric if the ratio between any two consecutive terms is always the same.
Now, we have to determine \[\dfrac{{{a_2}}}{{{a_1}}},\dfrac{{{a_3}}}{{{a_2}}},\dfrac{{{a_4}}}{{{a_3}}}\] we have
On substituting the corresponding values, we get
\[\dfrac{{{a_2}}}{{{a_1}}} = \dfrac{1}{{\sqrt 3 }}\]
\[\dfrac{{{a_3}}}{{{a_2}}} = \dfrac{3}{{\dfrac{1}{{\sqrt 3 }}}}\]
\[\dfrac{{{a_4}}}{{{a_3}}} = \dfrac{{\dfrac{1}{{3\sqrt 3 }}}}{3}\]
A progression of numbers is referred to as an arithmetic progression if there is always the same difference between any two subsequent terms.
Now, we have to also determine the values of \[{a_2} - {a_1},{a_3} - {a_2},{a_4} - {a_3}\]
Find \[{a_2} - {a_1}\]
On substituting the corresponding values, we get
\[{a_2} - {a_1} = \dfrac{1}{{\sqrt 3 }} - 1\]
On multiplying the denominator with the other term in the above expression, we get
\[ = \dfrac{{1 - \sqrt 3 }}{{\sqrt 3 }}\]
Find \[{a_3} - {a_2}\]
On substituting the corresponding values, we get
\[{a_3} - {a_2} = 3 - \dfrac{1}{{\sqrt 3 }}\]
On multiplying the denominator with the other term in the above expression, we get
\[ = \dfrac{{3\sqrt 3 - 1}}{{\sqrt 3 }}\]
Find \[{a_4} - {a_3}\]
On substituting the corresponding values, we get
\[{a_4} - {a_3} = \dfrac{1}{{3\sqrt 3 }} - 3\]
On multiplying the denominator with the other term in the above expression, we get
\[ = \dfrac{{1 - 9\sqrt 3 }}{{3\sqrt 3 }}\]
Now, from the previous calculations we have concluded that
\[\dfrac{{{a_2}}}{{{a_1}}} \ne \dfrac{{{a_3}}}{{{a_2}}} \ne \dfrac{{{a_4}}}{{{a_3}}}\]
And also, it is understood that
\[{a_2} - {a_1} \ne {a_3} - {a_2} \ne {a_4} - {a_3}\]
The series is hardly an A.P and neither a G.P because its terms cannot be expressed as \[a + (n - 1)d\] or \[a{r^{n - 1}}\]respectively.
Additionally, it is not an H.P because \[b = \dfrac{{2ac}}{{a + c}}\] in an H.P, which is not the case in this instance.
Therefore, the series \[1 + {3^{ - \dfrac{1}{2}}} + 3 + \dfrac{1}{{3\sqrt 3 }} + \ldots \] does not represent any of the given series.

Option ‘D’ is correct

Note: Students should be thorough with the formulas and concepts of the progressions like the formulas of A.P, G.P and H.P. These formulas play essential role in solving these types of problems. Applying wrong formulas will lead to wrong solution and hence consumes more time. So, one should be thorough with these concepts in order to get the desired solution.