Solution: $n^{th}$ term from the end is given by: $a_{n}=I(1 / r)^{n-1}$ where, $I$ is the last term, $r$ is the common ratio, $n$ is the $n^{th}$ term Given that, last term, $I=1 / 4374$...
Solution: Using the formula, $\begin{array}{l} \mathrm{T}_{n}=\mathrm{ar}^{\mathrm{n}-1} \\ \mathrm{a}=18 \\ \mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=(-12 / 18) \\ =-2 / 3 \\...
Solution: (i) $\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is $729 ?$ Using the formula, $\begin{array}{l} \mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1} \\ \mathrm{a}=\sqrt{3} \\ \mathrm{r}=\mathrm{t}_{2}...
Solution: (i) $\sqrt{2}, 1 / \sqrt{2}, 1 / 2 \sqrt{2}, 1 / 4 \sqrt{2}, \ldots$ is $1 / 512 \sqrt{2} ?$ Using the formula, $\begin{array}{l} T_{n}=a r^{n-1} \\ a=\sqrt{2} \\ r=t_{2} / t_{1}=(1 /...
Solution: Using the formula, $\mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}$ Given that, $\begin{array}{l} a=0.004 \\ r=t_{2} / t_{1}=(0.02 / 0.004) \\ =5 \\ T_{n}=12.5 \\ n=? \end{array}$...
term from the end of the G.P. 
.Solution: The $n^{th}$ term from the end is given by: $a_{n}=I(1 / r)^{n-1}$ where, $I$ is the last term, $r$ is the common ratio, $n$ is the nth term Given that, last term, $\mid=162$...
term of the G.P. 
term of the G.P. 
Solution: (i) nth term of the G.P. $\sqrt{3}, 1 / \sqrt{3}, 1 / 3 \sqrt{3}, \ldots$ It is known that, $t_{1}=a=\sqrt{3}, r=t_{2} / t_{1}=(1 / \sqrt{3}) / \sqrt{3}=1 /(\sqrt{3} \times \sqrt{3})=1 /...
term of the G.P. 
term of the G.P. 
Solution: (i) the $8^{\text {th }}$ term of the G.P., $0.3,0.06,0.012, \ldots$ It is known that, $\mathrm{t}_{1}=\mathrm{a}=0.3, \mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=0.06 / 0.3=0.2$ Using the...
term of the G.P. 
Solution: (i) the ninth term of the G.P. $1,4,16,64, \ldots$ It is known that, $t_{1}=a=1, r=t_{2} / t_{1}=4 / 1=4$ Using the formula. $\begin{array}{l}...
is a G.P.Solution: Given that, $a_{n}=2 / 3^{n}$ Consider $\mathrm{n}=1,2,3,4, \ldots$ since $\mathrm{n}$ is a natural number. Therefore, $\begin{array}{l} a_{1}=2 / 3 \\ a_{2}=2 / 3^{2}=2 / 9 \\ a_{3}=2 /...
Solution: (i) a, $3 \mathrm{a}^{2} / 4,9 \mathrm{a}^{3} / 16, \ldots$ Let $a=a, b=3 a^{2} / 4, c=9 a^{3} / 16$ In Geometric Progression, $\begin{array}{l} b^{2}=a c \\ \left(3 a^{2} / 4\right)^{2}=9...
Solution: (i) $4,-2,1,-1 / 2, \ldots$ Let $a=4, b=-2, c=1$ In $\mathrm{GP}$ $\begin{array}{l} b^{2}=a c \\ (-2)^{2}=4(1) \\ 4=4 \end{array}$ Therefore, the Common ratio $=r=-2 / 4=-1 / 2$ (ii) $-2 /...