Question

If \[{a_1},{a_2},{a_3},...\] are in AP then \[{a_p},{a_q},{a_r},...\] are in AP in \[p,q,r\] are in
A) AP
B) GP
C) HP
D) None of these

Answer 1

Hint: We know that an arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. The value by which consecutive terms increase or decrease is called the common difference.
Here, \[{a_1},{a_2},{a_3},...\] are in AP, so, \[{a_2} - {a_1} = {a_3} - {a_2} = ...\]
Using this formula, we can find a relation between \[{a_p},{a_q},{a_r},...\].
Simplifying this relation, we can find the relation between \[p,q,r\].

Complete step-by-step answer:
It is given that; \[{a_1},{a_2},{a_3},...\] are in AP and \[{a_p},{a_q},{a_r},...\] are in AP.
We have to find the relation between \[p,q,r\].
Since, \[{a_1},{a_2},{a_3},...\] are in AP.
So, the common differences are equal.
So, \[{a_2} - {a_1} = {a_3} - {a_2} = ...\]
Since, \[{a_p},{a_q},{a_r},...\] are in AP.
Let us consider, \[a\]is the initial term and \[d\] is the common difference. Then the \[n\] th element of the AP is \[{a_n} = a + (n - 1)d\]
Similarly, we have,
\[{a_p} = a + (p - 1)d\]
\[{a_q} = a + (q - 1)d\]
And, \[{a_r} = a + (r - 1)d\]
As, \[{a_p},{a_q},{a_r},...\] are in AP,
We have, \[{a_q} - {a_p} = {a_r} - {a_q}\]
Substitute the values we get,
$\Rightarrow$\[a + (q - 1)d - a - (p - 1)d = a + (r - 1)d - a - (q - 1)d\]
Simplifying we get,
$\Rightarrow$\[(q - 1 - p + 1)d = (r - 1 - q + 1)d\]
Simplifying again we get,
$\Rightarrow$\[q - p = r - q\]
So, we have, \[2q = p + r\]
We know that the series is in A.P if it is of the form \[2b = a + c\]
Hence, \[p,q,r\] are in AP.

$\therefore $ Option A is the correct answer.

Note: An itemized collection of elements in which repetitions of any sort are allowed is known as a sequence.
There are three types of sequence. A sequence in which every term is created by adding or subtracting a definite number to the preceding number is an arithmetic sequence.