
If \[a,b,c\]are in A.P. and \[|a|,|b|,|c| < 1\mid \]\[x = 1 + a + {a^2} + \ldots \ldots \infty \infty = 1 + b + {b^2} + \ldots \ldots \infty \left| {z = 1 + c + {c^2} \ldots \ldots \infty } \right|\] then \[x,y,z\] shall be in
A. A.P.
B. G.P.
C. H.P.
D. None of these
Answer
164.1k+ views
Hint:
Arithmetic Progression of a Series is a sequence of terms in which the difference between any two consecutive terms is always the same, according to the definition.
For instance, if \[x,y,z\] are in Arithmetic progression, then
\[y - x = z - y\]
Now, by interchanging the terms we get,
\[ \Rightarrow 2y = x + z\]
Therefore, if the given condition is satisfied, then the given sequence is in Arithmetic Progression. The condition to be satisfied is
\[2y = x + z\]
Formula used:
if \[x,y,z\] are in Arithmetic progression, then
\[y - x = z - y\]
\[2y = x + z\]
Sum of infinity of G.P
\[x = 1 + r + {r^2} + \ldots \ldots \infty = \frac{1}{{1 - r}}\]
Complete step-by-step solution:
It's assumed that a, b, and c are in Arithmetic Progression and \[|a|,|b|,|c| < 1\mid \]
We already know that, the sequence a, b, c is in Arithmetic progression,
We know that, it has been given
For \[x = 1 + a + {a^2} + \ldots \ldots \infty = \frac{1}{{1 - a}}\]---- (1)
For \[y = 1 + b + {b^2} + \ldots .\infty = \frac{1}{{1 - b}}\]---- (2)
For \[z = 1 + c + {c^2} + \ldots \ldots \infty = \frac{1}{{1 - c}}\]---- (3)
Since, a, b and c are in Arithmetic progression, \[1 - {\rm{a}},1 - {\rm{b}}\] and \[1 - {\rm{c}}\] are also in Arithmetic progression.
Therefore, \[\frac{1}{{1 - a}},\frac{1}{{1 - b}}\]and\[\frac{1}{{1 - c}}\] are in harmonic progression.
From the given information of equation (1), equation (2) and equation (3), it is given that
\[\frac{1}{{1 - a}} = x,\frac{1}{{1 - b}} = y,\frac{1}{{1 - c}} = z\]
Therefore, x, y and z are in Harmonic progression.
Therefore, if \[a,b,c\] are in A.P. and \[|a|,|b|,|c| < 1\], then x, y and z are in H.P.
Hence, the option C is correct.
Note:
If the preceding condition is not met, the terms are not in arithmetic progression. Then we must determine whether the terms are in geometric or harmonic progression. If the terms a, b, and c are in geometric progression, the equation \[{b^2} = ac\] should be satisfied.
Arithmetic Progression of a Series is a sequence of terms in which the difference between any two consecutive terms is always the same, according to the definition.
For instance, if \[x,y,z\] are in Arithmetic progression, then
\[y - x = z - y\]
Now, by interchanging the terms we get,
\[ \Rightarrow 2y = x + z\]
Therefore, if the given condition is satisfied, then the given sequence is in Arithmetic Progression. The condition to be satisfied is
\[2y = x + z\]
Formula used:
if \[x,y,z\] are in Arithmetic progression, then
\[y - x = z - y\]
\[2y = x + z\]
Sum of infinity of G.P
\[x = 1 + r + {r^2} + \ldots \ldots \infty = \frac{1}{{1 - r}}\]
Complete step-by-step solution:
It's assumed that a, b, and c are in Arithmetic Progression and \[|a|,|b|,|c| < 1\mid \]
We already know that, the sequence a, b, c is in Arithmetic progression,
We know that, it has been given
For \[x = 1 + a + {a^2} + \ldots \ldots \infty = \frac{1}{{1 - a}}\]---- (1)
For \[y = 1 + b + {b^2} + \ldots .\infty = \frac{1}{{1 - b}}\]---- (2)
For \[z = 1 + c + {c^2} + \ldots \ldots \infty = \frac{1}{{1 - c}}\]---- (3)
Since, a, b and c are in Arithmetic progression, \[1 - {\rm{a}},1 - {\rm{b}}\] and \[1 - {\rm{c}}\] are also in Arithmetic progression.
Therefore, \[\frac{1}{{1 - a}},\frac{1}{{1 - b}}\]and\[\frac{1}{{1 - c}}\] are in harmonic progression.
From the given information of equation (1), equation (2) and equation (3), it is given that
\[\frac{1}{{1 - a}} = x,\frac{1}{{1 - b}} = y,\frac{1}{{1 - c}} = z\]
Therefore, x, y and z are in Harmonic progression.
Therefore, if \[a,b,c\] are in A.P. and \[|a|,|b|,|c| < 1\], then x, y and z are in H.P.
Hence, the option C is correct.
Note:
If the preceding condition is not met, the terms are not in arithmetic progression. Then we must determine whether the terms are in geometric or harmonic progression. If the terms a, b, and c are in geometric progression, the equation \[{b^2} = ac\] should be satisfied.
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