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, $I=1 / 4374$...
Which term of the progression 18, -12, 8, … is 512/729 ?
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 \\...
Which term of the G.P.:
(i) √3, 3, 3√3, … is 729 ?
(ii) 1/3, 1/9, 1/27… is 1/19683 ?
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}...
Which term of the G.P.:
(i) √2, 1/√2, 1/2√2, 1/4√2, … is 1/512√2 ?
(ii) 2, 2√2, 4, … is 128 ?
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 /...
Which term of the progression 0.004, 0.02, 0.1, …. is 12.5?
Solution: Using the formula, $\mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}$ Given that, $\begin{array}{l} a=0.004 \\ r=t_{2} / t_{1}=(0.02 / 0.004) \\ =5 \\ T_{n}=12.5 \\ n=? \end{array}$...
Find the $4^{\text {th }}$ term from the end of the G.P. $2 / 27,2 / 9,2 / 3, \ldots ., 162$.
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, $\mid=162$...
Find:
(i) $n^{th}$ term of the G.P. $\sqrt{3}, 1 / \sqrt{3}, 1 / 3 \sqrt{3}, \ldots$
(ii) the $10^{\text {th }}$ term of the G.P. $\sqrt{2}, 1 / \sqrt{2}, 1 / 2 \sqrt{2}, \ldots .$
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 /...
Find:
(i) the $8^{\text {th }}$ term of the G.P. $0.3,0.06,0.012, \ldots .$
(ii) the $12^{\text {th }}$ term of the G.P. $1 / a^{3} x^{3}, a x, a^{5} x^{5}, \ldots .$
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...
Find:
(i) the ninth term of the G.P. $1,4,16,64, \ldots .$
(ii) the $10^{\text {th }}$ term of the G.P. $-3 / 4,1 / 2,-1 / 3,2 / 9, \ldots .$
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}...
Show that the sequence defined by $a_{n}=2 / 3^{n}, n \in N$ 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 /...
Show that each one of the following progressions is a G.P. Also, find the common ratio in each case:
(i) $a, 3 a^{2} / 4,9 a^{3} / 16, \ldots$
(ii) $1 / 2,1 / 3,2 / 9,4 / 27, \ldots$
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...
Show that each one of the following progressions is a G.P. Also, find the common ratio in each case:
(i) $4,-2,1,-1 / 2, \ldots$
(ii) $-2 / 3,-6,-54, \ldots .$
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 /...