A) HP
B) GP
C) AP
D) no particular order
Correct Answer: B
Solution :
The \[nth\]term of AP \[{{T}_{n}}=a+(n-1)d\] \[\therefore \] \[{{T}_{p}}=a+(p-1)d\] \[{{T}_{q}}=a(q-1)d\] \[{{T}_{r}}=a+(r-1)d\] and \[{{T}_{s}}=a+(s-1)d\] Since, \[{{T}_{p}},{{T}_{q}},{{T}_{r}}\]and \[{{T}_{s}}\]are in GP. \[\therefore \]Common ration, \[R=\frac{{{T}_{q}}}{{{T}_{p}}}=\frac{{{T}_{r}}}{{{T}_{q}}}\] \[=\frac{{{T}_{q}}-{{T}_{r}}}{{{T}_{p}}-{{T}_{q}}}\] \[=\frac{a+(q-1)d-[a+(r-1)d]}{a+(p-1)d-[a+(q-1)d]}\] \[=\frac{q-r}{p-q}\] ?(i) Similarly, \[R=\frac{r-s}{q-r}\] ?(ii) From Eqs. (i) and (ii), we get \[\frac{q-r}{p-q}=\frac{r-s}{q-r}\] \[\Rightarrow \] \[{{(q-r)}^{2}}=(r-s)(p-q)\] \[\therefore \] \[(p-q),(q-r)\]and \[(r-s)\]are in GP.
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