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8. Sequences and Series
medium
If $\frac{1}{{p + q}},\;\frac{1}{{r + p}},\;\frac{1}{{q + r}}$ are in $A.P.$, then
A
$p,\;,q,\;r$ are in $A.P.$
B
${p^2},\;{q^2},\;{r^2}$ are in $A.P.$
C
$\frac{1}{p},\;\frac{1}{q},\;\frac{1}{r}$ are in $A.P.$
D
None of these
Solution
(b) Since $\frac{1}{{p + q}},\;\frac{1}{{r + q}}$ and $\frac{1}{{q + r}}$ are in $A.P.$
$\therefore $ $\frac{1}{{r + q}} – \frac{1}{{p + q}} = \frac{1}{{q + r}} – \frac{1}{{r + p}}$
$ \Rightarrow $ $\frac{{p + q – r – p}}{{(r + p)(p + q)}} = \frac{{r + p – q – r}}{{(q + r)(r + p)}}$
$ \Rightarrow $ $\frac{{q – r}}{{p + q}} = \frac{{p – q}}{{q + r}}$ or ${q^2} – {r^2} = {p^2} – {q^2}$
$\therefore $ $2{q^2} = {r^2} + {p^2}$
Therefore ${p^2},\;{q^2},\;{r^2}$ are in $A.P.$
Standard 11
Mathematics
