Question

The series \[1{\text{ }},{\text{ }}1{\text{ }},{\text{ }}1{\text{ }}...\]is in
A.\[\left( 1 \right){\text{ }}A.P.\]
B. \[\left( 2 \right){\text{ }}G.P.\]
C.\[\left( 3 \right){\text{ }}both{\text{ }}\left( 1 \right){\text{ }}and{\text{ }}\left( 2 \right)\]
D.\[\;\left( 4 \right){\text{ }}none{\text{ }}of{\text{ }}these\]

Answer 1

Hint: We have to find the relation between the elements of the given series . We solve this question using the concept of arithmetic progression (A.P.) , geometric progression (G.P.) . We should also have the concept of the mean of the progressions and also the formula of $ {n^{(th)}} $ terms of A.P. and G.P. . On using the expression of A.P. and G.P we can compute that the given series is in which progression .

Complete step-by-step answer:
Given :
\[1{\text{ }},{\text{ }}1{\text{ }},{\text{ }}1{\text{ }} \ldots \ldots \ldots \]
To check the given series is in A.P.
From the given series :
First term \[\left( {{a_1}} \right){\text{ }} = {\text{ }}1\]
Second term \[\left( {{a_2}} \right){\text{ }} = {\text{ }}1\]
We know that the common difference \[\left( d \right)\]of the series is given as the difference of two consecutive terms :
So ,
Common difference \[\left( d \right){\text{ }} = {\text{ }}{a_2}{\text{ }} - {\text{ }}{a_1}\]
Common difference \[\left( d \right){\text{ }} = {\text{ }}1{\text{ }} - {\text{ }}1\]
Common difference \[\left( d \right){\text{ }} = {\text{ }}0\]
We know , that the formula of $ {n^{(th)}} $ term of A.P is given as :
\[{a_n}{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}\left( {n{\text{ }} - {\text{ }}1} \right){\text{ }} \times {\text{ }}d\]
Putting the values in the formula of $ {n^{(th)}} $ term , we get
\[{a_n}{\text{ }} = {\text{ }}1{\text{ }} + {\text{ }}\left( {n{\text{ }} - {\text{ }}1} \right){\text{ }} \times {\text{ }}0\]
\[{a_n}{\text{ }} = {\text{ }}1\]
Hence , we conclude that the series is in A.P. .
To check the given series is in G.P.
From the given series :
First term \[\left( {{a_1}} \right){\text{ }} = {\text{ }}1\]
Second term \[\left( {{a_2}} \right){\text{ }} = {\text{ }}1\]
We know that the common ratio (r) of the series is given as the ratio of the two consecutive terms of the series :
So ,
Common ratio \[\left( r \right){\text{ }} = {\text{ }}\dfrac{{{a_2}}}{{{a_1}}}\]
Common ratio \[\left( r \right){\text{ }} = {\text{ }}\dfrac{1}{1}\]
Common ratio \[\left( r \right){\text{ }} = {\text{ }}1\]
We know , that the formula of $ {n^{(th)}} $ term of G.P is given as :
 $ {a_n} = a \times {r^{(n - 1)}} $
Putting the values in the formula of $ {n^{(th)}} $ term , we get
 $ {a_n} = 1 \times {1^{(n - 1)}} $
\[{a_n}{\text{ }} = {\text{ }}1\]
Hence , we conclude that the series is in G.P. .
Thus , we conclude that the series is in both\[A.P\]. and\[G.P\].
Hence , the correct option is \[\left( 3 \right)\]
So, the correct answer is “Option 3”.

Note: The formula of mean of the three progression is given as :
(1) A.P.
Arithmetic mean \[ = {\text{ }}\dfrac{{\left( {{\text{ }}a{\text{ }} + {\text{ }}b{\text{ }}} \right){\text{ }}}}{2}\]
(2) G.P.
Geometric mean $ = \sqrt {(b \times c)} $
(3) H.P.
Harmonic mean \[ = {\text{ }}\dfrac{{2{\text{ }}ab{\text{ }}}}{{\left( {{\text{ }}a{\text{ }} + {\text{ }}b{\text{ }}} \right)}}\]