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

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Question

Given : $\frac{{{a_p} + {a_q}}}{2} = \frac{{{a_r} + {a_s}}}{2}$

$\begin{array}{l}
 \Rightarrow a + \left( {p - 1} \right)d + a + \left(

Vedclass · Accepted Answer

$r + s$

Vedclass · Answer

Given : $\frac{{{a_p} + {a_q}}}{2} = \frac{{{a_r} + {a_s}}}{2}$

$\begin{array}{l}
 \Rightarrow a + \left( {p - 1} ight)d + a + \left( {q - 1} ight)d\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = a + \left( {r - 1} ight)d + a + \left( {s - 1} ight)d
\end{array}$

$ \Rightarrow 2a + \left( {p + q} ight)d - 2d = 2a + \left( {r + s} ight)d - 2d$

$ \Rightarrow \left( {p + q} ight)d = \left( {r + s} ight)d \Rightarrow 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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