A) AP
B) GP
C) HP
D) None of the above series
Correct Answer: D
Solution :
[d] Given series is \[1+\frac{1}{\sqrt{3}}+3+\frac{1}{3\sqrt{3}}+.....\] Consider \[\frac{{{a}_{2}}}{{{a}_{1}}}=\frac{1}{\sqrt{3}},\] \[\frac{{{a}_{3}}}{{{a}_{2}}}=\frac{3}{1/\sqrt{3}},\] \[\frac{{{a}_{4}}}{{{a}_{3}}}=\frac{1/3\sqrt{3}}{3}\] Also find \[{{a}_{2}}-{{a}_{1}}=\frac{1}{\sqrt{3}}-1=\frac{1-\sqrt{3}}{\sqrt{3}}\] \[{{a}_{3}}-{{a}_{2}}=3-\frac{1}{\sqrt{3}}=\frac{3\sqrt{3}-1}{\sqrt{3}}\] \[{{a}_{4}}-{{a}_{3}}=\frac{1}{3\sqrt{3}}-3=\frac{1-9\sqrt{3}}{3\sqrt{3}}\] Thus, \[\frac{{{a}_{2}}}{{{a}_{1}}}\ne \frac{{{a}_{3}}}{{{a}_{2}}}\ne \frac{{{a}_{4}}}{{{a}_{3}}}\] and \[{{a}_{2}}-{{a}_{1}}\ne {{a}_{3}}-{{a}_{2}}\ne {{a}_{4}}-{{a}_{3}}\] Hence, given series is neither A.P, GP. nor HP.
You need to login to perform this action.
You will be redirected in
3 sec
