UPSC NDA Math Model Paper 3

Given series is, $1+3^{-\frac{1}{2}}+3+\frac{1}{3\sqrt{3}}\dots$ $\Rightarrow a_{1}=1, a_{2}=\frac{1}{\sqrt{3}}, a_{3}=3,$ $a_{4}=\frac{1}{3\sqrt{3}}\dots$ A sequence of numbers is called an arithmetic progression if the difference between any two consecutive terms is always same. $\Rightarrow a_{2}-a_{1}=\frac{1}{\sqrt{3}}-1=\frac{1-\sqrt{3}}{\sqrt{3}}$ Now, $a_{3}-a_{2}=3-\frac{1}{\sqrt{3}}=\frac{3\sqrt{3}-1}{\sqrt{3}}$ Here, $a_{2}-a_{1}\neq a_{3}-a_{2}$ Given series $1+3^{-\frac{1}{2}}+3+\frac{1}{3\sqrt{3}}\dots$ is not AP Now, A sequence of numbers is called a geometric progression if the ratio of any two consecutive terms is always same. $\frac{a_{2}}{a_{1}}=\frac{1}{\sqrt{3}}$ $\frac{a_{3}}{a_{2}}=\frac{3}{\frac{1}{\sqrt{3}}}=3\sqrt{3}$ Here, $\frac{a_{3}}{a_{2}}\neq\frac{a_{2}}{a_{1}}$ Given series $1+3^{-\frac{1}{2}}+3+\frac{1}{3\sqrt{3}}\dots$ is not GP A sequence of numbers is called a harmonic progression if the reciprocal of the terms are in AP. In simple terms, a,b,c,d,e,f are in HP if 1/a, 1/b, 1/c, 1/d, 1/e, 1/f are in AP. $\Rightarrow \frac{1}{a_{1}}=1, \frac{1}{a_{2}}=\sqrt{3}, \frac{1}{a_{3}}=\frac{1}{3}$ etc.. $\Rightarrow \frac{1}{a_{2}}-\frac{1}{a_{1}}=\sqrt{3}-1$ $\Rightarrow \frac{1}{a_{3}}-\frac{1}{a_{2}}=\frac{1}{3}-\sqrt{3}$ Here, $\frac{1}{a_{3}}-\frac{1}{a_{2}}\neq\frac{1}{a_{2}}-\frac{1}{a_{1}}$ Given series $1+3^{-\frac{1}{2}}+3+\frac{1}{3\sqrt{3}}\dots$ is not HP Hence, Given series $1+3^{-\frac{1}{2}}+3+\frac{1}{3\sqrt{3}}\dots$ is not AP, GP, HP