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8. Sequences and Series
hard
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
A
$r + s - 1$
B
$r + s - 2$
C
$r + s + 1$
D
$r + s$
(AIEEE-2012)
Solution
Given : $\frac{{{a_p} + {a_q}}}{2} = \frac{{{a_r} + {a_s}}}{2}$
$\begin{array}{l}
\Rightarrow a + \left( {p – 1} \right)d + a + \left( {q – 1} \right)d\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = a + \left( {r – 1} \right)d + a + \left( {s – 1} \right)d
\end{array}$
$ \Rightarrow 2a + \left( {p + q} \right)d – 2d = 2a + \left( {r + s} \right)d – 2d$
$ \Rightarrow \left( {p + q} \right)d = \left( {r + s} \right)d \Rightarrow p + q = r + s.$
Standard 11
Mathematics
