If the A.M. between p^{th} and q^{th} terms o | vedclass

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(D) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $p^{th}$ term is $a_p = a + (p - 1)d$.The $q^{th}$ term is $a_q = a +

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$r + s$

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(D) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $p^{th}$ term is $a_p = a + (p - 1)d$.The $q^{th}$ term is $a_q = a + (q - 1)d$.The $r^{th}$ term is $a_r = a + (r - 1)d$.The $s^{th}$ term is $a_s = a + (s - 1)d$.Given that the $A.M.$ of $a_p$ and $a_q$ is equal to the $A.M.$ of $a_r$ and $a_s$:$\frac{a_p + a_q}{2} = \frac{a_r + a_s}{2}$$\Rightarrow a_p + a_q = a_r + a_s$Substituting the values:$[a + (p - 1)d] + [a + (q - 1)d] = [a + (r - 1)d] + [a + (s - 1)d]$$2a + (p + q - 2)d = 2a + (r + s - 2)d$Subtracting $2a$ from both sides:$(p + q - 2)d = (r + s - 2)d$Assuming $d 
eq 0$,we can divide by $d$:$p + q - 2 = r + s - 2$Therefore,$p + q = r + s$.

If the $A.M.$ between $p^{th}$ and $q^{th}$ terms of an $A.P.$ is equal to the $A.M.$ between $r^{th}$ and $s^{th}$ terms of the same $A.P.$,then $p + q$ is equal to

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