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

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$$ Simplifying again we get, $\Rightarrowq - p = r - q$So, we have,$2q = p + r$We know that the series is in A.P if it is of the form$2b = a + c$Hence,$p,q,r$$ 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.