Question

If A.M between ${p^{th}}$and ${q^{th}}$ terms of an A.P be equal to the A.M between ${r^{th}}$ and ${s^{th}}$ terms of the A.P. then prove that $p + q$ is equal to
$A)r + s$
$B)\dfrac{{r - s}}{{r + s}}$
$C)\dfrac{{r + s}}{{r - s}}$
$D)r + s + 1$

Answer 1

Hint : First, we have to know about the arithmetic progression and arithmetic mean An arithmetic progression can be given by $a,(a + d),(a + 2d),(a + 3d),...$where $a$is the first term and $d$is a common difference.
$a,b,c$are said to be in arithmetic progression if the common difference between any two-consecutive number of the series is the same that is $b - a = c - b \Rightarrow 2b = a + c$
Where arithmetic means is all about finding the mean or average for the given terms, the sum of the given terms in the collection which divides the total count of the number.

Formula used:
The arithmetic progression can be expressed as ${a_n} = a + (n - 1)d$
Where $d$is the common difference, $a$is the first term, since we know that difference between consecutive terms is constant in any A.P
The arithmetic mean formula is $A.M = \dfrac{{\sum\limits_{i = 1}^n {{a_i}} }}{n}$, where n is the total count and ${a_i}$is the given data.

Complete step-by-step solution:
Since from the given that, the A.M between ${p^{th}}$and ${q^{th}}$terms of an A.P = the A.M between ${r^{th}}$and ${s^{th}}$terms of an A.P.
First, we find the common difference and first terms for each of the terms, ${p^{th}}$${q^{th}}$${r^{th}}$${s^{th}}$as below;
For the ${p^{th}}$ given A.P is of the form, ${a_n} = a + (n - 1)d \Rightarrow a + (p - 1)d$
Similarly, for the other terms we get, ${q^{th}}$the given A.P is of the form $a + (q - 1)d$
${r^{th}}$the given A.P is of the form $a + (r - 1)d$and ${s^{th}}$the given A.P is of the form $a + (s - 1)d$
Since from the given that, A.M of ${p^{th}}$and ${q^{th}}$terms are equals to = the A.M between ${r^{th}}$and ${s^{th}}$terms.
Now from the arithmetic mean formula $A.M = \dfrac{{\sum\limits_{i = 1}^n {{a_i}} }}{n}$and first apply this for ${p^{th}}$and ${q^{th}}$terms, where $\sum\limits_{i = 1}^n {{a_i}} = a + (p - 1)d + a + (q - 1)d$and $n = 2$(total quantity)
Hence the AM of ${p^{th}}$and ${q^{th}}$terms can be written as in the form of $\dfrac{{a + (p - 1)d + a + (q - 1)d}}{2}$
Again, from the arithmetic mean formula $A.M = \dfrac{{\sum\limits_{i = 1}^n {{a_i}} }}{n}$and apply this for ${r^{th}}$and ${s^{th}}$terms, where $\sum\limits_{i = 1}^n {{a_i}} = a + (r - 1)d + a + (s - 1)d$and $n = 2$(total quantity)
Similarly, the AM of ${r^{th}}$and ${s^{th}}$terms can be written as in the form of $\dfrac{{a + (r - 1)d + a + (s - 1)d}}{2}$
Where $2$is the total count and ${a_i}$is the given data from the A.P
Hence these two terms are equals as per the given question, then we get;
$\dfrac{{a + (p - 1)d + a + (q - 1)d}}{2} = \dfrac{{a + (r - 1)d + a + (s - 1)d}}{2}$
Calling the common terms; we get $2a + (p - 1)d + (q - 1)d = 2a + (r - 1)d + (s - 1)d$
$ \Rightarrow (p - 1)d + (q - 1)d = (r - 1)d + (s - 1)d$
$ \Rightarrow (p + q)d - 2d = (r + s)d - 2d$
$ \Rightarrow (p + q)d = (r + s)d$
$ \Rightarrow (p + q) = (r + s)$
Hence after solving the equal terms, we get $ \Rightarrow (p + q) = (r + s)$
Hence option $A)r + s$ is correct.

Note: For the AP, the number of terms can be represented as
3 terms; $(a - d),a,(a + d)$
The resulting sequence also will be in AP. In an arithmetic progression, the sum of terms from beginning and end will be constant.
In the A.M $\sum\limits_{i = 1}^n {{a_i}} $is the sum of the given terms.
AP is the arithmetic progression and AM is the arithmetic mean which is the average of the sum of the AP divides the total count.