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If the A.M. between $p^{\text {th }}$ and $q^{\text {th }}$ terms of an A.P. is equal to the A.M. between $r^{\text {th }}$ and $s^{\text {th }}$ terms of the same A.P., then $p+q$ is equal to
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Verified Answer
The correct answer is:
$r+s$
$r+s$
Given: $\frac{a_p+a_q}{2}=\frac{a_r+a_s}{2}$
$$
\begin{aligned}
& \Rightarrow a+(p-1) d+a+(q-1) d \\
& \quad=a+(r-1) d+a+(s-1) d \\
& \Rightarrow 2 a+(p+q) d-2 d=2 a+(r+s) d-2 d \\
& \Rightarrow(p+q) d=(r+s) d \Rightarrow p+q=r+s .
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow a+(p-1) d+a+(q-1) d \\
& \quad=a+(r-1) d+a+(s-1) d \\
& \Rightarrow 2 a+(p+q) d-2 d=2 a+(r+s) d-2 d \\
& \Rightarrow(p+q) d=(r+s) d \Rightarrow p+q=r+s .
\end{aligned}
$$
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