Question

If the first and \[\left( 2n-1 \right){{~}^{th}}\] terms of an A.P, G.P and H.P are equal and their \[{{n}^{th}}\] terms are respectively a, b and c then always
A) \[a\text{ }=\text{ }b\text{ }=\text{ }c\]
B) \[a\ge ~b~\ge c\]
C) \[a+\text{ }c\text{ }=\text{ }b\]
D) \[ac-\text{ }{{b}^{2}}~=\text{ }0\]

Answer 1

Hint:
Find the value of \[{{n}^{th}}\]term in terms of first and \[\left( 2n-1 \right){{~}^{th}}\]term for AP, GP and HP series using the formulas of the \[{{n}^{th}}\] term by letting the common difference: d and common ratio: r, for the respective series. Then try to relate them to get the answer to this question.

Formula Used:
For A.P series we know that,
\[{{n}^{th}}\]= first term + (n-1) common difference

Now, for GP series we know that the \[{{n}^{th}}\]term
\[=\text{ }first\text{ }term~{{\left( common\text{ }ratio \right)}^{n-1}}\]

Complete step by step solution:
It is given that the first and \[\left( 2n-1 \right){{~}^{th}}\]terms of AP, GP and HP are equal.
Let the first term be x and \[\left( 2n-1 \right){{~}^{th}}\]term be y
For A.P series we know that,
\[{{n}^{th}}\]= first term + (n-1) common difference
The first term is x and \[\left( 2n-1 \right){{~}^{th}}\]term is y
And let the common difference = d
\[\Rightarrow \]y= x + (2n-2) d
The \[{{n}^{th}}\]= a\[=~x+\text{ }\left( n-1 \right)\text{ }d\] (1)
{it is given that \[{{n}^{th}}\]term of AP series =a }
 Now,
The sum of first term and \[\left( 2n-1 \right){{~}^{th}}\]term
= x+ y
= x+ x + (2n-2) d
= 2x + (2n-2) d
= 2[x+(n-1)d]
From (1)
= 2[a]
\[\Rightarrow \]x+ y = 2[a]
\[\Rightarrow \] (x+ y)/2 = a
Or,
\[\Rightarrow \] a = (x+ y)/2 (2)

Now, for GP series we know that the \[{{n}^{th}}\] term
= first term \[\times \] (common ratio)\[^{n-1}\]
Let the common ratio = r
\[\Rightarrow \] \[{{n}^{th}}\] term\[\begin{align}
  & =\text{ }x~{{r}^{n-1}}~=\text{ }b \\
 & ~ \\
\end{align}\] (3)
        {it is given in that the \[{{n}^{th}}\] term of GP series =b}
Now,
 (2n-1) $^{th}$ term = \[x~{{r}^{2n-2}}\]
\[\Rightarrow \]first term \[\times \] (2n-1) $^{th}$ term = \[x\times ~x\times ~{{r}^{2n-2}}\]= \[{{x}^{2}}\times {{r}^{2n-2}}\]
= \[{{(x\times {{r}^{n-1}})}^{2}}\]
From (3)
\[\Rightarrow x~\times y\text{ }=\text{ }b{{~}^{2}}\]
        Taking square root on both sides
We get,
\[\sqrt{xy}=b\] (4)
Now, for HP series,
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 1/a, 1/b, 1/c, 1/d, 1/e, 1/f are in AP.
For two terms that is x and y
x= first term,
y= \[\left( 2n-1 \right){{~}^{th}}\]term
Harmonic Mean, c = (2 x y) / (x + y)
                                                                                     (5)
  From (2), (4) and (5) the conditions satisfied are
\[\begin{align}
  & a\text{ }\ge ~b\text{ }\ge \text{ }c \\
 & a\text{ }b\text{ }=\text{ }{{c}^{2}} \\
 & \Rightarrow ab-{{c}^{2}}~=\text{ }0 \\
\end{align}\]
Hence,

B) and C) are correct.

Additional Information:
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.
Arithmetic Progression (AP): A sequence of numbers is called an arithmetic progression if the difference between any two consecutive terms is always the same.
If ‘a’ is the first term and d is the common difference,
nth term of an AP = 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.
If ‘a’ is the first term and ‘r’ is the common ratio,
nth term of a GP= a rn-1
Sum of ‘n’ terms of a GP(r<1) \[=\text{ }\left[ a\text{ }\left( 1\text{ }\text{ }{{r}^{n}} \right)\left] \text{ }/\text{ } \right[1\text{ }\text{ }r \right]\]
Sum of ‘n’ terms of a GP (r > 1)\[=\text{ }\left[ a\text{ }\left( {{r}^{n}}~\text{ }1 \right)\left] \text{ }/\text{ } \right[r\text{ }\text{ }1 \right]\]
Sum of infinite terms of a GP (r < 1) = (a) / (1 – r)
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 1/a, 1/b, 1/c, 1/d, 1/e, 1/f are in AP.
For two terms ‘a’ and ‘b’,
Harmonic Mean = (2 a b) / (a + b)
For two numbers, if A, G and H are respectively the arithmetic, geometric and harmonic means, then
A ≥ G ≥ H
A H = G2, i.e., A, G, H are in GP

Note:
The knowledge of progressions series that is AP, GP and HP series that is formula for \[{{n}^{th}}\] term of these series which are
If ‘a’ is the first term and d is the common difference, \[{{n}^{th}}\]term of an AP = a + (n-1) d
If ‘a’ is the first term and ‘r’ is the common ratio, \[{{n}^{th}}\]term of a GP = a rn-1 is important for students to answer this question.