A) \[\frac{r-p}{q-p}\]
B) \[\frac{q-p}{r-q}\]
C) \[\frac{r-q}{q-p}\]
D) 1
Correct Answer: C
Solution :
[c] : Let common difference be d. \[{{a}_{p}}={{a}_{1}}+(p-1)d,{{a}_{q}}={{a}_{1}}+(q-1)d,\] \[{{a}_{r}}={{a}_{1}}+(r-1)d\] As \[{{a}_{p}},{{a}_{q}},{{a}_{r}}\] are in G.R \[\therefore \]\[\frac{{{a}_{q}}}{{{a}_{p}}}=\frac{{{a}_{r}}}{{{a}_{q}}}=\frac{{{a}_{q}}-{{a}_{r}}}{{{a}_{p}}-{{a}_{q}}}\](by law of proportions) or\[\frac{{{a}_{q}}}{{{a}_{p}}}=\frac{{{a}_{r}}}{{{a}_{q}}}=\frac{{{a}_{1}}+(q-1)d-{{a}_{1}}-(r-1)d}{{{a}_{1}}+(p-1)d-{{a}_{1}}-(q-1)d}=\frac{q-r}{p-q}\] or\[\frac{{{a}_{q}}}{{{a}_{p}}}=\frac{q-r}{p-q}=\frac{r-q}{q-p}\]
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