A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: B
Solution :
[b] Let the 3n terms of GP. be \[a,ar,a{{r}^{2}},.....a{{r}^{n-1}},a{{r}^{n}},a{{r}^{n+1}},...a{{r}^{2n-1}},a{{r}^{2n}},a{{r}^{2n+1}},.....,\]\[a{{r}^{3n-1}}\]. Then \[{{S}_{1}}=a+ar+a{{r}^{2}}+......+a{{r}^{n-1}}=\frac{a(1-{{r}^{n}})}{1-r}\] \[{{S}_{2}}=a{{r}^{n}}+a{{r}^{n+1}}+.....+a{{r}^{2n-1}}=\frac{a{{r}^{n}}(1-{{r}^{n}})}{1-r}\] \[{{S}_{3}}=a{{r}^{2n}}+a{{r}^{2n+1}}+.....+a{{r}^{3n-1}}=\frac{a{{r}^{2n}}(1-{{r}^{n}})}{1-r}\] Clearly \[\frac{{{S}_{2}}}{{{S}_{1}}}=\frac{{{S}_{3}}}{{{S}_{2}}}={{r}^{n}}\]You need to login to perform this action.
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