If \[{p^{th}},{q^{th}},{r^{th}}\] and \[{s^{th}}\] terms of an A.P. be in G.P., then \[(p - q),(q - r),(r - s)\] will be in
A. G.P.
B. A.P.
C. H.P.
D. None of these
Answer
54.6k+ views
Hint:
We know that, \[{T_n} = a + \left( {n-1} \right)d\] is the nth term of an arithmetic progression (A.P) with first term 'a' and common difference- ’d’. The common difference can be positive, negative, or zero. And we can determine the nature of AP based on the value of the common difference d.
Complete step-by-step solution:
We have been given in the question, that
The \[{p^{th}},{q^{th}},{r^{th}}\] and \[{s^{th}}\] terms of an A.P. be in G.P
Let A be the first term D be the common difference of A.P.
We know that,
In terms of p-th term, we have
\[{a_p} = a + (p - 1)d\]
In terms of q-th term, we have
\[{a_q} = a + (q - 1)d\]
In terms of r-th term, we have
\[{a_r} = a + (r - 1)dn\]
In terms of s-th term, we have
\[{a_s} = a + (s - 1)dn\]
It is given that \[{a_p},{a_q},{a_r}\] and \[{a_s}\] is in Geometric progression.
Substitute all the values obtained from the above calculations, we obtain
Therefore, \[\frac{{{a_q}}}{{{a_p}}} = \frac{{{a_r}}}{{{a_q}}} = \frac{{{a_q} - {a_r}}}{{{a_p} - {a_q}}} = \frac{{q - r}}{{p - q}} \ldots \ldots \](1)
Now, we have to cancel the similar terms to simplify:
\[\frac{{{a_r}}}{{{a_q}}} = \frac{{{a_s}}}{{{a_r}}} = \frac{{{a_r} - {a_s}}}{{{a_q} - {a_r}}} = \frac{{r - s}}{{q - r}} \ldots \ldots \](2)
On simplifying, from equations (1) and (2), we get
\[\frac{{q - r}}{{p - q}} = \frac{{r - s}}{{q - r}}\]
Therefore, \[p - q,q - r\] and \[r - s\] are in Geometric progression.
Hence, the option A is correct.
Note:
The assumption that a common difference can never be negative or always positive is incorrect. So, student should determine whether the given terms have a mathematical formula to represent them; if so, it is a progression; otherwise, it is a sequence. However, a mathematical formula should be used to represent the given terms in progression. Identify the common difference between increasing and decreasing arithmetic progressions or determine whether a given sequence is increasing or decreasing.
We know that, \[{T_n} = a + \left( {n-1} \right)d\] is the nth term of an arithmetic progression (A.P) with first term 'a' and common difference- ’d’. The common difference can be positive, negative, or zero. And we can determine the nature of AP based on the value of the common difference d.
Complete step-by-step solution:
We have been given in the question, that
The \[{p^{th}},{q^{th}},{r^{th}}\] and \[{s^{th}}\] terms of an A.P. be in G.P
Let A be the first term D be the common difference of A.P.
We know that,
In terms of p-th term, we have
\[{a_p} = a + (p - 1)d\]
In terms of q-th term, we have
\[{a_q} = a + (q - 1)d\]
In terms of r-th term, we have
\[{a_r} = a + (r - 1)dn\]
In terms of s-th term, we have
\[{a_s} = a + (s - 1)dn\]
It is given that \[{a_p},{a_q},{a_r}\] and \[{a_s}\] is in Geometric progression.
Substitute all the values obtained from the above calculations, we obtain
Therefore, \[\frac{{{a_q}}}{{{a_p}}} = \frac{{{a_r}}}{{{a_q}}} = \frac{{{a_q} - {a_r}}}{{{a_p} - {a_q}}} = \frac{{q - r}}{{p - q}} \ldots \ldots \](1)
Now, we have to cancel the similar terms to simplify:
\[\frac{{{a_r}}}{{{a_q}}} = \frac{{{a_s}}}{{{a_r}}} = \frac{{{a_r} - {a_s}}}{{{a_q} - {a_r}}} = \frac{{r - s}}{{q - r}} \ldots \ldots \](2)
On simplifying, from equations (1) and (2), we get
\[\frac{{q - r}}{{p - q}} = \frac{{r - s}}{{q - r}}\]
Therefore, \[p - q,q - r\] and \[r - s\] are in Geometric progression.
Hence, the option A is correct.
Note:
The assumption that a common difference can never be negative or always positive is incorrect. So, student should determine whether the given terms have a mathematical formula to represent them; if so, it is a progression; otherwise, it is a sequence. However, a mathematical formula should be used to represent the given terms in progression. Identify the common difference between increasing and decreasing arithmetic progressions or determine whether a given sequence is increasing or decreasing.
Last updated date: 03rd Jun 2023
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Total views: 54.6k
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