Question 13 Marks
Find the sum of the following series:
0.6 + 0.66 + 0.666 + ... to n terms.
Answer$0.6+0.66+0.666+\&\ ...\text{ to n}$
$=6\times0.1+6\times0.11+6\times0.111+\ \dots$
$=\frac69\Big\{\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ \dots\ +-\Big\}$
$=\frac69\Big\{\Big(1-\frac{1}{10}\Big)+\Big(1-\frac{1}{100}\Big)+\ \dots\ +\Big\}$
$=\frac{6}{9}\Big\{\text{n}-\Big(\frac{1}{10}+\frac{1}{10^2}+\ \dots\ +\frac{1}{10^\text{n}}\Big)\Big\}$
$=\frac69\Bigg[\text{n}-\frac{1}{10}\frac{\big\{1-\big(\frac{1}{10}\big)^\text{n}\big\}}{\big(1-\frac{1}{10}\big)}\Bigg]$
$=\frac{6}{9}\Big[\text{n}-\frac19\Big(1-\frac{1}{10^\text{n}}\Big)\Big]$
View full question & answer→Question 23 Marks
If $\frac{\text{a}+\text{bx}}{\text{a}-\text{bx}}=\frac{\text{b}+\text{cx}}{\text{b}-\text{cx}}=\frac{\text{c}+\text{dx}}{\text{c}-\text{dx}}(\text{x}\neq0),$ then show that a, b, c and d are in G.P.
Answer$\frac{\text{a}+\text{bx}}{\text{a}-\text{bx}}=\frac{\text{b}+\text{cx}}{\text{b}-\text{cx}}=\frac{\text{c}+\text{dx}}{\text{c}-\text{dx}}$
Now, $\frac{\text{a}+\text{bx}}{\text{a}-\text{bx}}=\frac{\text{b}+\text{cx}}{\text{b}-\text{cx}}$
Applying componendo and dividendo
$\Rightarrow\frac{(\text{a}+\text{bx})+(\text{a}-\text{bx})}{(\text{a}+\text{bx})-(\text{a}-\text{bx})}=\frac{(\text{b}+\text{cx})+(\text{b}-\text{cx})}{(\text{b}+\text{cx})-(\text{b}-\text{cx})}$
$\Rightarrow\frac{2\text{a}}{2\text{bx}}=\frac{2\text{b}}{2\text{cx}}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{\text{b}}{\text{c}}$
Similiarly, $\frac{(\text{b}+\text{cx})+(\text{b}-\text{cx})}{(\text{b}+\text{cx})-(\text{b}-\text{cx})}=\frac{(\text{c}+\text{dx})+(\text{c}-\text{dx})}{(\text{c}+\text{dx})-(\text{b}-\text{dx})}$
$\Rightarrow\frac{\text{b}}{\text{c}}=\frac{\text{c}}{\text{d}}$
$\therefore$ a, b, c and d are in G.P.
View full question & answer→Question 33 Marks
Which term of the G.P.:$\sqrt{3},3,3\sqrt{3},\ \dots\text{is}729?$
Answer$\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\frac{1}{4\sqrt{2}},\ \dots\text{is}\frac{1}{512\sqrt{2}}?$$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{a}=\sqrt{2},\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac12$
$\text{t}_\text{n}=\frac{1}{512\sqrt{2}},\text{n}=?$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\frac{1}{5125\sqrt{2}}=\big(\sqrt{2}\big)\Big(\frac12\Big)^{\text{n}-1}$
$\frac{1}{512\times\sqrt{2}\times\sqrt{2}}=\Big(\frac{1}{2}\Big)^{\text{n}-1}$
$\frac{1}{1024}=\Big(\frac12\Big)^{\text{n}-1}$
$\Big(\frac12\Big)^{10}=\Big(\frac12\Big)^{{\text{n}-1}}$
$10=(\text{n}-1)$
$\text{n}=11$
$\therefore\text{term is 11}^\text{th}.$
$\sqrt{3},3,3\sqrt{3},\ \dots\text{is}729?$
$\text{a}=\sqrt{3},\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=,\text{n}=?,\text{t}_\text{n}=729$
Now,
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$729=\big(\sqrt{3}\big)(\text{r})^{\text{n}-1}$
Now,
$\text{r}=\frac{\text{t}_2}{\text{t}_1}=\frac{3}{\sqrt{3}}=\sqrt{3}$
$729=\big(\sqrt{3}\big)\big(\sqrt{3}\big)^{\text{n}-1}$
$729=\Big(\sqrt{3}\Big)^\text{n}$
$(3)^6=\big(\sqrt{3}\big)^{\text{n}}$
$\big(\sqrt{3}\big)^{12}=\big(\sqrt{3}\big)^{\text{n}}$
$\Rightarrow{\text{n}=12}$
$\therefore12^{\text{th}}\text{ term is 729}.$
View full question & answer→Question 43 Marks
If S denotes the sum of an infinite G.P. and $S_1$ denotes the sum of the squares of its terms, then prove that the first terms and common ratio are respectively $\frac{2\text{SS}_1}{\text{S}^2+\text{S}_1}\text{ and }\frac{\text{S}^2-\text{S}_1}{\text{S}^2+\text{S}_1}.$
Answer$\text{S}=\text{a}+\text{ar}+\text{ar}^2+\text{ar}^3+\ \dots$
$\text{S}=\frac{\text{a}}{1-\text{r}}\ \cdots(1)$
$\text{S}_1=\text{a}^2+\text{a}^2\text{r}^2+\text{a}^2\text{r}^4+\text{a}^2\text{r}^6\ \dots$
$\text{S}_1=\frac{\text{a}^2}{1-\text{r}^2}\ \dots(2)$
$\text{S}^2=\frac{\text{a}^2}{(1-\text{r})^2}$
$\text{S}^2=\frac{\text{S}_1\big(1-\text{r}^2\big)}{\big(1-\text{r}^2\big)}$
$\big(1-\text{r}\big)\text{S}^2=\text{S}_1(1+\text{r})$
$\text{S}^2-\text{S}^2\text{r}=\text{S}_1-\text{S}_1\text{r}$
$\text{S}_1\text{r}+\text{S}^2\text{r}=\text{S}^2-\text{S}_1$
$\text{r}=\frac{\text{S}^2-\text{S}_1}{\text{S}_1+\text{S}^2}$
Put r in equition (1)
$\text{S}(1-\text{r})=\text{a}$
$\text{a}=\text{S}\bigg[1-\frac{\text{S}^2-\text{S}_1}{\text{S}^2+\text{S}_1}\bigg]$
$\text{a}=\text{S}\bigg[\frac{\text{S}^2+\text{S}_1-\text{S}^2+\text{S}_1}{\text{S}^2+\text{S}_1}\bigg]$
$\text{a}=\frac{2\text{S}\text{S}_1}{\text{S}^2+\text{S}_1}$
View full question & answer→Question 53 Marks
Find the sum of the following geometric series:$\frac{3}{5}+\frac{4}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+\ ...\ \text{to 2n terms;}$
AnswerThe series can be written as:
$2\Big(\frac15+\frac{1}{5^3}+\frac{1}{5^5}+\ ...\ \text{n terms}\Big)+4\Big(\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+\ ...\ \text{n terms}\Big)$
For the first part $\text{a}=\frac15$ and the common ratio $\text{r}=\frac{1}{5^2}=\frac{1}{25}$
thus the sum is:
$3\Big(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+\ ...\ \text{n terms}\Big)=3\cdot\frac{\frac{1}{5}\Big(1-\Big(\frac{1}{25}\Big)^\text{n}\Big)}{1-\frac{1}{25}}$
$=\frac{5}{8}\Big(1-\frac{1}{5^{2\text{n}}}\Big)$
For the second part $\text{a}=\frac{1}{25}$ and common ratio $\text{r}=\frac{1}{25}$ then
$4\Big(\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+\ ...\ \text{n terms}\Big)=4\cdot\frac{\frac{1}{25}\Big(1-\Big(\frac{1}{25}\Big)^\text{n}\Big)}{1-\frac{1}{25}}$
$=\frac{1}{6}\Big(1-\frac{1}{5^{2\text{n}}}\Big)$
Thus the sum is:
$\frac35+\frac{4}{5^2}+\frac{3}{5^3}+\ ...\ 2\text{n terms}=\frac{5}{8}\Big(1-\frac{1}{5^{2\text{n}}}\Big)+\frac16\Big(1-\frac{1}{5^{2\text{n}}}\Big)$
View full question & answer→Question 63 Marks
How many terms of the sequence $\sqrt{3},3,3\sqrt{3},\ ...$ must be taken to make the sum $39+13\sqrt{3}?$
Answer$\sqrt{3},3,3\sqrt{3},\ ...$
$\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}$
$\text{a}=\sqrt{3},\text{r}=\frac{3}{\sqrt{3}}=\sqrt{3},\text{S}_\text{n}=39+13\sqrt{3}$
Putting into formula
$39+13\sqrt{3}=\frac{\sqrt{3}\big(\big(\sqrt{3}\big)^\text{n}-1\big)}{\sqrt{3}-1}$
$39+13\sqrt{3}=\frac{\big(\sqrt{3}\big)^{\text{n}-1}-\sqrt{3}}{\sqrt{3}-1}$
$\big(39+13\sqrt{3}\big)\big(\sqrt{3}-1\big)=\big(\sqrt{3}\big)^{\text{n}+1}-\sqrt{3}$
$39\sqrt{3}-39+39-13\sqrt{3}=\big(\sqrt{3}\big)^{\text{n}+1}-\sqrt{3}$
$26\sqrt{3}+\sqrt{3}=\big(\sqrt{3}\big)^{\text{n}+1}$
$\big(27\sqrt{3}\big)^1=\big(\sqrt{3}\big)^{\text{n}+1}$
$\big(\sqrt{3}\big)^6\big(\sqrt{3}\big)^1=\big(\sqrt{3}\big)^{\text{n}+1}$
$7=\text{n}+1$
$\Rightarrow\text{n}=6$
View full question & answer→Question 73 Marks
Find the sum of the following geometric series:
$\sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}+\ ... \text{ to $8$ terms;}$
AnswerHere the first term of the series is $\text{a}=\sqrt{2}$ and the common ratio is $\text{r}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}$Thus the sum of the G.P. up to $8^{th}$ terms is:
$\text{S}_8\frac{\text{a}(1-\text{r}^8)}{1-\text{r}}=\frac{\sqrt{2}\Big(1-\big(\frac12\big)^8\Big)}{1-\frac12}$
$=2\sqrt{2}\Big(1-\frac{1}{256}\Big)=\frac{255\sqrt{2}}{128}$
View full question & answer→Question 83 Marks
Insert 5 geometric means between $16$ and $\frac{1}{4}.$
Answer$5$ Geometric means between $16$ and $\frac{1}{4}.$
Let $G_1, G_2, G_3, G_4, G_5$ be $5$ geometric means between $a = 16$ and $\text{b}=\frac{1}{4}$
Then, $16, G_1, G_2, G_3, G_4, G_5$ $\frac{1}{4}$ is a G.P. with $a = 16$, $\text{b}=\frac{1}{4}$
$\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{\text{n}+1}}$
$=\Bigg(\frac{\frac{1}{4}}{16}\Bigg)^{\frac{1}{5+1}}=\Big(\frac{1}{26}\Big)^{\frac{1}{6}}=\Big(\frac{1}{2}\Big)$
$\therefore\text{G}_1=\text{ar}=16\times\Big(\frac12\Big)=8$
$\text{G}_2=\text{ar}^2=16\times\frac{1}{4}=4$
$\text{G}_3=\text{ar}^3=16\times\frac{1}{8}=2$
$\text{G}_4=\text{ar}^4=16\times\frac{1}{16}=1$
$\text{G}_5=\text{ar}^5=16\times\frac{1}{{2}^5}=\frac{1}{2}$
Hence, $8,4,2,1,\frac{1}{2}$ are $5$ geometric means between $16$ and $\frac{1}{4}.$
View full question & answer→Question 93 Marks
The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio $(3+2\sqrt{2}):\big(3-2\sqrt{2}\big).$
AnswerA.M. between two numbers a and b (a > b) is $\frac{\text{a}+\text{b}}{2}$
Also, Geometric means between 2 numbers is $\sqrt{\text{ab}}$
A.M. = 2G.M
$\Rightarrow\frac{\text{a}+\text{b}}{2}=2\sqrt{\text{ab}}$
$\text{a}+\text{b}=4\sqrt{\text{ab}}$
$\frac{\text{a}+\text{b}}{2\sqrt{\text{ab}}}=\frac{2}{1}$
$\frac{\text{a}+\text{b}+2\sqrt{\text{ab}}}{\text{a}+\text{b}-2\sqrt{\text{ab}}}=\frac{2+1}{2-1}=\frac{3}{1}$ [By componendo and dividendo]
$\frac{\big(\sqrt{\text{a}}+\sqrt{\text{b}}\big)^2}{\big(\sqrt{\text{a}}-\sqrt{\text{b}}\big)^2}=\frac{\big(\sqrt{3}\big)^2}{(1)^2}$
$\frac{\sqrt{\text{a}}+\sqrt{\text{b}}}{\sqrt{\text{a}}-\sqrt{\text{b}}}=\frac{\sqrt{3}}{1}$
By componendo and dividendo, we get
$\frac{\big(\sqrt{\text{a}}+\sqrt{\text{b}}\big)+\big(\sqrt{\text{a}}-\sqrt{\text{b}}\big)}{\big(\sqrt{\text{a}}+\sqrt{\text{b}}\big)-\big(\sqrt{\text{a}}-\sqrt{\text{b}}\big)}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$\frac{\sqrt{\text{a}}}{\sqrt{\text{b}}}=\frac{\big(\sqrt{3}+1\big)^2}{\big(\sqrt{3}-1\big)^2}=\frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}$
$=\frac{4+2\sqrt{3}}{4-2\sqrt{3}}$
$\frac{\text{a}}{\text{b}}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Thus, $\text{a}:\text{b}\big(2+\sqrt{3}\big):\big(2-\sqrt{3}\big).$
View full question & answer→Question 103 Marks
Find the sum of the following series:
7 + 77 + 77 + ... to n terms.
Answer$7 + 77 + 777 + ...\text{ to n terms.}=7[1+11+111+\text{to n terms}]$
$=\frac79[9 + 99+999+\ ... \ \text{n terms}]$
$=\frac{7}{9}\Big[\big(10-1\big)+\big(10^2-1\big)+\big(10^3-1\big)+\dots\text{n terms}\Big]$
$=\frac{7}{9}\Big[\big(10+10^2+10^3+\dots\text{n terms}\big)\Big]-\frac{7}{9}(1+1+1+\ \dots\text{to n terms})$
$=\frac79\cdot\frac{10(10^\text{n}-1)}{10-1}-\frac{7\text{n}}{9}$
$=\frac{7}{81}(10^{\text{n}+1}-9\text{n}-10)$
View full question & answer→Question 113 Marks
The common ratio of a G.P. is 3 and the last term is 486. If the sum of these terms be 728, Find the first term.
Answerr = 3, last term is 486
Sum of terms = Sn = 728, a = ₹
We know that
$\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}$
$728=\frac{\text{a}(3^\text{n}-1)}{3-1}$
Also, $\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{t}_\text{n}=486$
$486=\text{a}(3)^{\text{n}-1}$
$\text{a}(3^{\text{n}-1})=3^5\times2$
$\text{n}=6$
And, $\Rightarrow\text{a}=2$
View full question & answer→Question 123 Marks
Find the sum of the following geometric progrssions:$(\text{a}^2-\text{b}^2),(\text{a}-\text{b}),\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big),\ ...\text{to n terms}$
Answer$(\text{a}^2-\text{b}^2),(\text{a}-\text{b}),\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big), ...\text{n terms}$$\text{a}=\text{a}^2-\text{b}^2,\text{r}=\frac{\text{a}-\text{b}}{\text{a}^2-\text{b}^2}=\frac{1}{\text{a}+\text{b}},\text{n}=\text{n}$
$\text{S}_\text{n}=\text{a}\frac{(\text{r}^{1-\text{n}})}{\text{r}-1}$ $[\because\text{r}<1]$
$\text{S}_\text{n}=(\text{a}^2-\text{b}^2)\frac{\big(1-\frac{1}{(\text{a}+\text{b})^\text{n}}\big)}{1-\frac{1}{\text{a}+\text{b}}}$
$=\frac{(\text{a}-\text{b})((\text{a}+\text{b})^\text{n}-1)}{\frac{(\text{a}+\text{b})^{-1}(\text{a}+\text{b})^\text{n}(\text{a}+\text{b})-1}{(\text{a}+\text{b})}}$
$=\frac{\text{a}-\text{b}}{{(\text{a}+\text{b})^{\text{n}-2}}}\Big\{\frac{(\text{a}+\text{b})^\text{n}-1}{(\text{a}+\text{b})-1}\Big\}$
View full question & answer→Question 133 Marks
Find the sum of the following geometric series:
0.15 + 0.015 + 0.0015 + ... to 8 terms;
AnswerHere, a = 0.15 and $\text{r}=\frac{\text{a}_2}{\text{a}_1}=\frac{0.015}{0.15}=\frac{1}{10}.$$\text{S}_8=\text{a}\Big(\frac{1-\text{r}^{8}}{1-\text{r}}\Big)$
$=0.15\Bigg(\frac{1-\big(\frac{1}{10}\big)^{8}}{1-\frac{1}{10}}\Bigg)$
$=0.15\Bigg(\frac{1-\frac{1}{{10}^8}}{\frac{1}{10}}\Bigg)$
$=\frac16\Big(1-\frac{1}{{10^8}}\Big)$
View full question & answer→Question 143 Marks
Find the sum of the following geometric series:$\frac{\text{a}}{1+\text{i}}+\frac{\text{a}}{(1+\text{i})^2}+\frac{\text{a}}{(1+\text{i})^3}+\ ...\ +\frac{\text{a}}{(1+\text{i})^\text{n}}.$
Answer$\frac{\text{a}}{1+\text{i}}+\frac{\text{a}}{(1+\text{i})^2}+\frac{\text{a}}{(1+\text{i})^3}+\ ...\ +\frac{\text{a}}{(1+\text{i})^\text{n}}.$ $\text{a}=\frac{\text{a}}{1+\text{i}},\text{r}=\frac{\frac{\text{a}}{(1+\text{i})^2}}{\frac{\text{a}}{1+\text{i}}}=\frac{1}{1+\text{i}}$ $\text{S}_\text{n}=\text{a}\frac{(1-\text{r}^{\text{n}})}{1-\text{r}}$ $=\frac{\text{a}}{1+\text{i}}\times\frac{\Big(1-\Big(\frac{1}{1+\text{i}}\Big)^\text{n}\Big)}{1-\frac{1}{1+\text{i}}}$ $=\frac{\text{a}}{1+\text{i}}\times\frac{1+\text{i}}{(-\text{i})}(1-(1+\text{i})^\text{n})$$=-\text{ai}(1-(1+\text{i})^{-\text{n}})$
View full question & answer→Question 153 Marks
Find the rational number having the following decimal expansions:$0.\overline{231}$
Answer$0.\overline{231}=0.231231231\ \dots$$=0.231+0.000231+0.000000231+\ $
$=\frac{231}{10^3}+\frac{231}{10^6}+\frac{231}{10^9}+\ \dots$
$=\frac{231}{10^3}\Big(1+\frac{1}{10^3}+\frac{1}{10^6}+\ \dots\Big)$
$=\frac{231}{1000}\Bigg(\frac{1}{1-\frac{1}{1000}}\Bigg)$
$0.\overline{231}=\frac{231}{999}$
View full question & answer→Question 163 Marks
If a, b, c, d and p are different real numbers such that:
$(\text{a}^2+\text{b}^2+\text{c}^2)\text{p}^2-2(\text{ab}+\text{bc}+\text{cd})\text{p}+(\text{b}^2+\text{c}^2+\text{a}^2)\le0,$ then show that a, b, c and d are in G.P.
Answer$(\text{a}^2+\text{b}^2+\text{c}^2)\text{p}^2-2(\text{ab}+\text{bc}+\text{cd})\text{p}+(\text{b}^2+\text{c}^2+\text{a}^2)\le0,$
$\Rightarrow(\text{a}^2\text{p}^2+\text{b}^2\text{p}^2+\text{c}^2\text{p}^2)-2(\text{a}\text{b}\text{p}+\text{b}\text{c}\text{p}+\text{c}\text{d}\text{p})+(\text{b}^2+\text{c}^2+\text{d}^2)\le0$
$\Rightarrow(\text{a}^2\text{p}^2-2\text{a}\text{b}\text{p}+\text{b}^2)+(\text{b}^2\text{p}^2-2\text{b}\text{c}\text{p}+\text{c}^2)+(\text{c}^2\text{p}^2-2\text{c}\text{d}\text{p}+\text{d}^2)\le0$
$\Rightarrow(\text{a}\text{p}-\text{b})^2+(\text{b}\text{p}-\text{c})^2+(\text{c}\text{p}-\text{d})^2\le0$
$\Rightarrow(\text{a}\text{p}-\text{b})^2+(\text{b}\text{p}-\text{c})^2+(\text{c}\text{p}-\text{d})^2=0$
$\Rightarrow(\text{a}\text{p}-\text{b})^2=0$
$\Rightarrow\text{p}=\frac{\text{b}}{\text{a}}$
Also, $(\text{bp}-\text{c})^2=0$
$\Rightarrow\text{p}=\frac{\text{c}}{\text{b}}$
Similiarly, $\Rightarrow(\text{cp}-\text{d})^2=0$
$\Rightarrow\text{p}=\frac{\text{d}}{\text{c}}$
$\therefore\frac{\text{b}}{\text{a}}=\frac{\text{c}}{\text{b}}=\frac{\text{d}}{\text{c}}$
Thus, a, b, c and d are in G.P.
View full question & answer→Question 173 Marks
Find an infinite G.P. whose first term is 1 and each term is sum of all the terms which Follow it.
Answer$\text{a}=1$
$\text{a}_\text{n}=\text{a}_{\text{n}+1}+\text{a}_{\text{n}+2}+\text{a}_{\text{n}+3}+\ \dots$
$\text{ar}^{\text{n}-1}=\text{ar}^\text{n}+\text{ar}^{\text{n}+1}+\text{ar}^{\text{n}+2}+\dots$
$\text{ar}^{\text{n}-1}=\text{ar}^{\text{n}}\big(1+\text{r}+\text{r}^2+\dots\infty)$
$1=\text{r}\Big(\frac{1}{1-\text{r}}\Big)$
$1-\text{r}=\text{r}$
$1=2\text{r}$
$\text{r}=\frac12$
$\text{G.P. is }1,\frac{1}{2},\frac{1}{4},\frac{1}{8}\dots$
View full question & answer→Question 183 Marks
Find the sum of the following series to infinit:$\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+\ ...\infty$
AnswerThe G.P can be written as follows:
$\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+\ ...\infty$
$=\Big(\frac13+\frac{1}{3^3}+\frac{1}{3^5}+\cdots\infty\Big)+\Big(\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+\cdots\infty\Big)$
$=\frac{\frac{1}{3}}{1-\frac{1}{3^2}}+\frac{\frac{1}{5^2}}{1-\frac{1}{5^2}}$
$=\frac{3}{8}+\frac{1}{24}$
$=\frac{10}{24}$
$=\frac{5}{12}$
View full question & answer→Question 193 Marks
Evaluate the following:$\sum_\limits{\text{n}=2}^{10}(4^\text{n})$
Answer$\sum_\limits{\text{n}=2}^{10}(4^\text{n})$
$=4^2+4^3+4^4+\ \dots\ +4^{10}$
$\text{a}=4^2,\text{r}=\frac{4^3}{4}=4,\text{n}=9$
$\text{S}_{10}=\frac{\text{a}(\text{r}^9-1)}{1-\text{r}}$
$=\frac{4^2(4^9-1)}{4-1}$
$=\frac13\big[4^{11}-16\big]$
$=\frac{16}{3}\big[4^{9}-1\big]$
View full question & answer→Question 203 Marks
Which term of the progression $18,-12,8,\dots\text{is}\frac{512}{729}?$
Answer$18,-12,8,\dots\text{is}\frac{512}{729}$
$\text{a}=18,\text{n}=?,\text{t}_\text{n}=\frac{512}{729},\text{r}=\frac{\text{t}_{\text{n}-1}}{\text{t}_\text{n}}$
$\text{r}=\frac{\text{t}_2}{\text{t}_1}=\frac{-12}{18}=\frac{-2}{3}$
Also,
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\frac{512}{729}=(18)\Big(\frac{-2}{3}\Big)^{\text{n}-1}$
$\frac{2^9}{36}\times\frac{1}{2\times3^2}=\Big(\frac{-2}{3}\Big)^{\text{n}-1}$
$\Big(\frac23\Big)^8=(-1)^{\text{n}-1}\Big(\frac23\Big)^{\text{n}-1}$
$\text{n}=9$
View full question & answer→Question 213 Marks
Find the sum of the following geometric series:
$\sqrt{7},\sqrt{21},3\sqrt{7},\dots\text{to n terms}$
AnswerHere the first term of the G.P. is $\text{a}=\sqrt{7}$ and the common ratio is $\text{r}=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$
Thus the sum of the G.P. is:
$\sqrt{7}+\sqrt{21}+3\sqrt{7}+\cdots\text{to n terms}=\frac{\sqrt{7}\Big(\big(\sqrt{3}\big)^\text{n}-1\Big)}{\sqrt{3}-1}=\frac{\sqrt{7}\Big(3^{\frac{\text{n}}{2}}-1\Big)}{\sqrt{3}-1}$
View full question & answer→Question 223 Marks
If $\frac{1}{\text{a}+\text{b}},\frac{1}{2\text{b}},\frac{1}{\text{b}+\text{c}}$ are three consecutive terms ofan A.P., prove that a, b, c are the three consecutive terms of a G.P.
Answer$\frac{1}{\text{a}+\text{b}},\frac{1}{2\text{b}},\frac{1}{\text{b}+\text{c}}\text{ are in A.P.}$
$\frac{2}{2\text{b}}=\frac{1}{(\text{a}+\text{b})}+\frac{1}{(\text{b}+\text{c})}$
$\frac{1}{\text{b}}=\frac{\text{b}+\text{c}+\text{a}+\text{b}}{(\text{a}+\text{b})(\text{b}+\text{c})}$
$\frac{1}{\text{b}}=\frac{2\text{b}+\text{c}+\text{a}}{\text{ab}+\text{ac}+\text{b}^2+\text{bc}}$
$\text{ab}+\text{ac}+\text{b}^2+\text{bc}=2\text{b}^2+\text{bc}+\text{ba}$
$\text{b}^2+\text{ac}=2\text{b}^2$
$\text{b}^2=\text{ac}$
So,
a, b, c are in G.P.
View full question & answer→Question 233 Marks
Which term of the G.P.:$\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\frac{1}{4\sqrt{2}},\ \dots\text{is}\frac{1}{512\sqrt{2}}?$
Answer$\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\frac{1}{4\sqrt{2}},\ \dots\text{is}\frac{1}{512\sqrt{2}}?$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{a}=\sqrt{2},\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac12$
$\text{t}_\text{n}=\frac{1}{512\sqrt{2}},\text{n}=?$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\frac{1}{5125\sqrt{2}}=\big(\sqrt{2}\big)\Big(\frac12\Big)^{\text{n}-1}$
$\frac{1}{512\times\sqrt{2}\times\sqrt{2}}=\Big(\frac{1}{2}\Big)^{\text{n}-1}$
$\frac{1}{1024}=\Big(\frac12\Big)^{\text{n}-1}$
$\Big(\frac12\Big)^{10}=\Big(\frac12\Big)^{{\text{n}-1}}$
$10=(\text{n}-1)$
$\text{n}=11$
$\therefore\text{term is 11}^\text{th}.$
Here, first term, $\text{a}=\sqrt{2}$
and common ratio, $\text{r}=\frac12$
Let the $n^{th}$ term be $\frac{1}{512\sqrt{2}}.$
$\therefore\text{a}_\text{n}=\frac{1}{512\sqrt{2}}$
$\Rightarrow\text{a}\text{r}^{\text{n}-1}=\frac{1}{512\sqrt{2}}$
$\Rightarrow\big(\sqrt{2}\big)\Big(\frac12\Big)^{\text{n}-1}=\frac{1}{512\sqrt{2}}$
$\Rightarrow\Big(\frac12\Big)^{\text{n}-1}=\frac{1}{1024}$
$\Rightarrow\Big(\frac12\Big)^{\text{n}-1}=\Big(\frac12\Big)^{10}$
$\Rightarrow\text{n}-1=10$
$\Rightarrow\text{n}=11$
Thus, the $11^{th}$ term of the given G.P. is $\frac{1}{512\sqrt{2}}.$
View full question & answer→Question 243 Marks
Which term of the G.P.:$2,2\sqrt{2},4,\ \dots\text{is}128?$
Answer$\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\frac{1}{4\sqrt{2}},\ \dots\text{is}\frac{1}{512\sqrt{2}}?$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{a}=\sqrt{2},\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac12$
$\text{t}_\text{n}=\frac{1}{512\sqrt{2}},\text{n}=?$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\frac{1}{5125\sqrt{2}}=\big(\sqrt{2}\big)\Big(\frac12\Big)^{\text{n}-1}$
$\frac{1}{512\times\sqrt{2}\times\sqrt{2}}=\Big(\frac{1}{2}\Big)^{\text{n}-1}$
$\frac{1}{1024}=\Big(\frac12\Big)^{\text{n}-1}$
$\Big(\frac12\Big)^{10}=\Big(\frac12\Big)^{{\text{n}-1}}$
$10=(\text{n}-1)$
$\text{n}=11$
$\therefore\text{term is 11}^\text{th}.$
$2,2\sqrt{2},4,\ \dots\text{is}128?$
$\text{a}=2,\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{2\sqrt{2}}{2}=\sqrt{2},\text{n}=?$
$\text{t}_\text{n}=128$
Also,
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$128=(2)\big(\sqrt{2}\big)^{\text{n}-1}$
$\frac{128}{2}=\big(\sqrt{2}\big)^{\text{n}-1}$
$64=\big(\sqrt{2}\big)^{\text{n}-1}$
$(2)^6=\big(\sqrt{2}\big)^{\text{n}-1}$
$\Rightarrow12={\text{n}-1}$
${\text{n}-13}$
$\therefore13^{\text{th}}\text{ term is 128}.$
View full question & answer→Question 253 Marks
Cunstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.
AnswerLet the roots of the quadratic equation be a and b.
$\text{A}=\frac{\text{a}+\text{b}}{2}$
$\therefore\text{a}+\text{b}=2\text{A}\cdots(\text{i})$
Also, $\text{G}^2=\text{ab}\cdots(\text{ii})$
The quadratic equation having roots a and b is given by $\text{x}^2 - (\text{a} + \text{b})\text{x} + \text{ab} = 0.$
$\therefore\text{x}^2-2\text{Ax}+\text{G}^2=0$ [Using (i) and (ii)]
View full question & answer→Question 263 Marks
If a, b, c are in A.P. and a, b, d are in G.P., show that a, (a - b), (d - c) are in G.P.
Answera, b, c are in A.P.
$\therefore2\text{b}=\text{a}+\text{c }\cdots(\text{i})$
Also, a, b, d are in G.P.
$\therefore\text{b}^2=\text{ad }\cdots{\text{(ii}})$
Now,
$(\text{a}-\text{b})^2=\text{a}^2-2\text{ab}+\text{b}^2$
$\Rightarrow(\text{a}-\text{b})^2=\text{a}^2-\text{a}(\text{a}+\text{c})+\text{ad}$ [Using (i) and (ii)]
$\Rightarrow(\text{a}-\text{b})^2=\text{a}^2-\text{a}^2-\text{ac}+\text{ad}$
$\Rightarrow(\text{a}-\text{b})^2=\text{ad}-\text{ac}$
$\Rightarrow(\text{a}-\text{b})^2=\text{a}(\text{d}-\text{c})$
Therefore, a, (a - b) and (d - c) are in G.P.
View full question & answer→Question 273 Marks
Which term of the G.P.:$\frac13,\frac19,\frac{1}{17}\ ...\text{is}\frac{1}{19683}?$
Answer$\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\frac{1}{4\sqrt{2}},\ \dots\text{is}\frac{1}{512\sqrt{2}}?$$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{a}=\sqrt{2},\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac12$
$\text{t}_\text{n}=\frac{1}{512\sqrt{2}},\text{n}=?$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\frac{1}{5125\sqrt{2}}=\big(\sqrt{2}\big)\Big(\frac12\Big)^{\text{n}-1}$
$\frac{1}{512\times\sqrt{2}\times\sqrt{2}}=\Big(\frac{1}{2}\Big)^{\text{n}-1}$
$\frac{1}{1024}=\Big(\frac12\Big)^{\text{n}-1}$
$\Big(\frac12\Big)^{10}=\Big(\frac12\Big)^{{\text{n}-1}}$
$10=(\text{n}-1)$
$\text{n}=11$
$\therefore\text{term is 11}^\text{th}.$
$\frac13,\frac19,\frac{1}{17}\ ...\text{is}\frac{1}{19683}?$
$\text{a}=\frac{1}{3},\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3},\text{t}_\text{n}=\frac{1}{19683},\text{n}=?$
Now,
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\frac{1}{19683}=\Big(\frac13\Big)\Big(\frac13\Big)^{\text{n}-1}=\Big(\frac13\Big)^\text{n}$
$\Big(\frac13\Big)^9=\Big(\frac13\Big)^\text{n}$
$\Rightarrow{\text{n}=9}$
$\therefore9^{\text{th}}\text{ term of G.P. is}\frac{1}{19683}.$
View full question & answer→Question 283 Marks
If a, b, c are in G.P., prove that:
$\text{a}(\text{b}^2+\text{c}^2)=\text{c}(\text{a}^2+\text{b}^2)$
Answera, b and c are in G.P.
$\therefore\text{b}^2=\text{ac}\ \cdots(1)$
$\text{L.H.S}=\text{a}(\text{b}^2+\text{c}^2)$
$={\text{ab}^2}{+\text{ac}^2}$
$=\text{a}(\text{ac})+\text{c}\big(\text{b}^2\big)$ [Using (1)]
$=\text{c}\big(\text{a}^2+\text{b}^2\big)$
$=\text{R.H.S}$
View full question & answer→Question 293 Marks
Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is equal to $\frac{32}{81}.$
AnswerHere,
a = 4
$\text{A}_3-\text{a}_5=\frac{31}{81}$
$\text{ar}^2-\text{ar}^4=\frac{32}{81}$
$\text{r}^24\big(1-\text{r}^2\big)=\frac{32}{81}$
$\text{r}^2\big(1-\text{r}^2\big)=\frac{8}{81}$
Let $\text{r}^2=\text{A}$
$\text{A}(1-\text{A})=\frac{8}{81}$
$\text{A}-\text{A}^2=\frac{8}{81}$
$81\text{A}-81\text{A}^2=8$
$81\text{A}^2-81\text{A}+8=0$
$\text{A}=\frac{81\pm\sqrt{(81)^2-4\times81\times8}}{81\times2}$
$=\frac{81\pm\sqrt{6561-2592}}{162}$
$=\frac{81\pm\sqrt{3969}}{162}$
$=\frac{81\pm63}{162}$
$=\frac{81\pm63}{162}\text{ or }\frac{81-63}{162}$
$=\frac{144}{162}\text{ or }\frac{18}{162}$
$\text{r}^2=\frac89\text{ or }\frac19$
$\text{r}=\pm\frac{2\sqrt{2}}{3}\text{ or } \pm\frac13$
Since it is a decreasing G.P.
$\text{r}=\frac{2\sqrt{2}}{3},\frac{1}{3}$
$\text{S}_\infty=\frac{4}{1-\frac{2\sqrt{2}}{3}}\text{ and }\text{S}_\infty=\frac{4}{1-\frac13}$
$\text{S}_\infty=\frac{12}{3-2\sqrt{2}},6$
View full question & answer→Question 303 Marks
Prove that: $\Big(2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots\infty\Big)=2$
Answer$2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots\infty$
$=2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots\infty$
$=\Big(\frac14+\frac18+\frac{3}{16}+\frac{4}{32}+\ \dots\infty\Big)$
$=2$
$=2^{5}\cdots(1)$
$\text{S}=\frac{1}{4}+\frac28+\frac{3}{16}+\frac{4}{32}+\ \dots\infty$
$\text{S}=\Big(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\dots\infty\Big)2$
$\frac{\text{S}}{2}=\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\dots\infty$
$=\frac{\frac{1}{4}}{1-\frac12}$
$=\frac{1}{4}\times\frac21$
$\text{S}=\frac12$
$\text{S}=1$
Thus, $2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots=2^1=2$
View full question & answer→Question 313 Marks
If A.M. and G.M. of two positive numbers a and b are 10 and 8 respectively, find the numbers.
AnswerLet the two of equation be a and b.
AM = 10
$\therefore\ \frac{\text{a}+\text{b}}{2}=10$
$\Rightarrow\text{a}+\text{b}=20\ \cdots(\text{i})$
Also, G = 8
$\therefore\sqrt{\text{ab}}=8$
$\Rightarrow\text{ab}=8^2$
$\Rightarrow\text{ab}=64\ \cdots(\text{ii})$
Using (i) and (ii):
$\Rightarrow\text{a}(20-\text{a})=64$
$\Rightarrow\text{a}^2-20\text{a}+64=0$
$\Rightarrow\text{a}^2-16\text{a}-4\text{a}+64=0$
$\Rightarrow\text{a}(\text{a}-16)-4(\text{a}-16)=0$
$\Rightarrow(\text{a}-16)(\text{a}-4)=0$
$\Rightarrow\text{a}=4,16$
If a = 4, then b = 16.
And, if a = 16, then b = 4.
View full question & answer→Question 323 Marks
Find the sum of the following series:
5 + 55 + 555 + ... to n terms.
Answer5 + 55 + 555 + ... to n terms.
Taking 5 common from each term.
5[1 + 11 + 111 + ... n terms]
Dividing and multiplying by 9
$=\frac59[9 + 99+999+\ ... \ \text{n terms}]$
$=\frac{5}{9}\Big[\big(10-1\big)+\big(10^2-1\big)+\big(10^3-1\big)+\dots\text{n terms}\Big]$
$=\frac{5}{9}\Big[\big(10+10^2+10^3+\dots\text{n terms}\big)-\text{n}\Big]\text{this is G.P.}$
So,
$\text{S}_\text{n}=\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}$
$\text{a}=10, \text{r}=10,\text{n}=\text{n}$
$=\frac59\Big[\frac{10(10^\text{n}-1)}{10-1}-\text{n}\Big]$
$=\frac{5}{9\times9}\big(10^{\text{n}+1}-10-9\text{n})$
$=\frac{5}{81}(10^{\text{n}+1}-9\text{n}-10)$
View full question & answer→Question 333 Marks
Find the geometric means of the following pairs of numbers:
$a^3b$ and $ab^3$
Answer$a^3b$ and $ab^3$
Geometric means between a and $\text{b}=\sqrt{\text{ab}}$
$a = a^3b, b = ab^3$
$\therefore\text{Geometric means}=\sqrt{\text{a}^3\text{b}\times\text{ab}^3}$
$=\sqrt{\text{a}^4\text{b}^4}=\text{a}^2\text{b}^2$
View full question & answer→Question 343 Marks
Find the rational number having the following decimal expansions:$0.\overline{3}$
Answer$0.\overline{3}=0.3333\ \dots$
$=0.3+0.03+0.003+\ \dots$
$=\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+\ \dots$
$=\frac{3}{10}\Big(1+\frac{1}{10}+\frac{1}{10^2}+\ \dots\Big)$
$=\frac{3}{10}\Bigg(\frac{1}{1-\frac{1}{10}}\Bigg)$
$=\frac{3}{10}\times\frac{10}{9}$
$=\frac39$
$0.\overline{3}=\frac13$
View full question & answer→Question 353 Marks
If the G.P.'s $5, 10, 20, ... $ and $1280, 640, 320, ...$ have their $n^{th}$ terms equal, find the value of $n$.
Answer$5, 10, 20, ... n$ term
$1280, 640, 320, ..., n$ terms.
Let $t_n$ be the general term if first G.P. and $t_n'$ be general term of recourd G.P. whose term are equal.
a for first G.P. $= 5$
a for second G.P. $= 1280$
r for first G.P $=\frac{10}{5}=2$
r = for second G.P $\frac{640}{1280}=\frac{1}{2}$
$t_n = ar^{n-1}$
Applying and equating for both $G.P_n'$
$(5) (2)^{\text{n}-1} = 1280\Big(\frac{1}{2}\Big)^{\text{n}-1}$
$(2)^{\text{n}-1}=\frac{1280}{5}\Big(\frac12\Big)^{\text{n}-1}$
$=256\Big(\frac12\Big)^{\text{n}-1}$
$=2^8\Big(\frac12\Big)^{\text{n}-1}$
$\frac{(2)^{\text{n}-1}}{28}=\Big(\frac12\Big)^{\text{n}-1}=2^{\text{n}-1}=2^{-\text{n}+1}$
$\Rightarrow2\text{n}=10$
$\text{n}=5$
View full question & answer→Question 363 Marks
Find the sum of the following geometric series:
$\frac12-\frac13+\frac12-\frac34+\ ...\text{to 5 terms;}$
Answer$\frac12-\frac13+\frac12-\frac34+\ ...\text{to 5 terms;}$
$\text{a}=\frac{2}{9},\text{r}=\frac{\frac{-1}{3}}{\frac{2}{9}}=\frac{-1}{3}\times\frac92=\frac{-3}{2},\text{n}=5$
$\text{S}_5=\text{a}\frac{(1-\text{r}^5)}{1-\text{r}}$
$=\frac{2}{9}\cdot\frac{\Big(1-\big(\frac{-3}{2}\big)^5\Big)}{1-\big(\frac{-3}{2}\big)}$
$=\frac{2}{9}\cdot\frac{\Big(1+\frac{243}{32}\Big)}{1+\frac{3}{2}}$
$=\frac{2}{9}\cdot\frac{(275)}{32}\times\frac{2}{5}$
$=\frac{55}{72}$
View full question & answer→Question 373 Marks
Prove that: $\Big(9^{\frac{1}{3}}.9^{\frac19}.9^{\frac{1}{27}}\dots\infty\Big)=3.$
Answer$9^{\frac{1}{3}}\times9^{\frac{1}{9}}\times9^{\frac{1}{27}}\dots\infty$
$=9^{\big(\frac13+\frac19+\frac{1}{27}+\ \dots\big)}$
$=9^{\Bigg(\frac{\frac13}{1-\frac13}\Bigg)}$ $\Big[\text{Using }\text{S}_\infty=\frac{\text{a}}{1-\text{r}}\Big]$
$=9^{\big(\frac{1}{3}\times\frac32\big)}$
$=9^\frac{1}{2}$
$=3$
So,
$9^{\frac{1}{3}}\times9^{\frac{1}{9}}\times9^{\frac{1}{27}}\dots\infty=3$
View full question & answer→Question 383 Marks
How many terms of the G.P. $3,\frac{3}{2},\frac{3}{4},\ \dots$ be taken together to make $\frac{3069}{512}?$
AnswerHere,
$3,\frac{3}{2},\frac{3}{4},\ \dots\text{ is a G.P.}$
And $\text{S}_\text{n}=\frac{3069}{512},\text{a}=3,\text{r}=\frac{1}{2}$
$\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}$
$\frac{3069}{512}=\frac{3\Big(1-\big(\frac12\big)^\text{n}\Big)}{1-\frac12}$
$\frac{3069}{512}=\frac{3(2^\text{n}-1)}{2^\text{n}\times\frac12}$
$\frac{1023}{512}=\frac{2(2^\text{n}-1)}{2^\text{n}}$
$10232^\text{n}=1024.2^\text{n}-1024$
$1024=2^\text{n}$
$\Rightarrow2^{10}=2^\text{n}$
$\Rightarrow\text{n}=10$
View full question & answer→Question 393 Marks
Find the geometric means of the following pairs of numbers:2 and 8
Answer2 and 8
Geometric means between a and $\text{b}=\sqrt{\text{ab}}\cdots$
Here, a = 2, b = 8
$\therefore\text{Geometric means}=\sqrt{2\times8}$
$=\sqrt{16}=4$
View full question & answer→Question 403 Marks
Find the sum of the following series:
0.5 + 0.55 + 0.555 + ... to n terms.
Answer$0.5 + 0.55 + 0.555 + ...\text{ to n terms.}$
$=5\times0.1+5\times0.11+5\times0.111+\ ...$
$=\frac59\Big\{\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\dots\ +-\Big\}$
$=\frac59\Big\{\Big(1-\frac{1}{10}\Big)+\Big(1-\frac{1}{100}\Big)+\ \dots\ +\Big\}$
$=\frac59\Big\{\text{n}-\Big(\frac{1}{10}+\frac{1}{10^2}+\ \dots\ +\frac{1}{10^\text{n}}\Big)\Big\}$
$=\frac59\Bigg[\text{n}-\frac{1}{10}\frac{\big\{1-\big(\frac{1}{10}\big)^\text{n}\big\}}{\big(1-\frac{1}{10}\big)}\Bigg]$
$=\frac59\Big[\text{n}\frac{1}{9}\Big(1-\frac{1}{10^\text{n}}\Big)\Big]$
View full question & answer→Question 413 Marks
If $p^{th}, q^{th}, r^{th}$ and sth terms of an A.P. be in then prove that $p - q, q - r, r - s$ are in $G.P$.
AnswerHere, Let R be common ratio,
$a_p, a_q, a_r, a_s$_ of AP are in G.P.
$\text{R}=\frac{\text{a}_\text{q}-\text{a}_\text{r}}{\text{a}_\text{p}-\text{a}_\text{q}}\text{ (Ratio property)}$
$=\frac{[\text{a}+(\text{q}-1)\text{d}]-[\text{a}+(\text{r}-1)\text{d}]}{[\text{a}+(\text{p}-1)\text{d}]-[\text{a}+(\text{q}-1)\text{d}]}$
$=\frac{(\text{q}-\text{r})\text{d}}{(\text{p}-\text{q})\text{d}}$
$\text{R}=\frac{\text{q}-\text{r}}{\text{p}-\text{q}}\cdots(1)$
Now,
$\text{R}=\frac{\text{a}_\text{r}}{\text{a}_\text{q}}=\frac{\text{a}_\text{s}}{\text{a}_\text{r}}$
$=\frac{\text{a}_\text{r}}{\text{a}_\text{q}}=\frac{\text{a}_\text{s}}{\text{a}_\text{r}}\text{ (Ratio property})$
$=\frac{[\text{a}+(\text{r}-1)\text{d}]-[\text{a}+(\text{s}-1)\text{d}]}{[\text{a}+(\text{q}-1)\text{d}]-[\text{a}+(\text{r}-1)\text{d}]}$
$=\frac{(\text{r}-\text{s})\text{d}}{(\text{q}-\text{r})\text{d}}$
$\text{R}=\frac{\text{r}-\text{s}}{\text{q}-\text{r}}\cdots(2)$
From equation as (1) and (2)
$\frac{\text{q}-\text{r}}{\text{p}-\text{q}}=\frac{\text{r}-\text{s}}{\text{p}-\text{r}}$
$\Rightarrow(\text{p}-\text{q}),(\text{q}-\text{r}),(\text{r}-\text{s})\text{ are in G.P.}$
View full question & answer→Question 423 Marks
The sum of three numbers $a, b, c$ in A.P. is $18$. If a and b are each increased by $4$ and c is increased by $36$, the new numbers from a G.P.
Find $a, b, c$.
AnswerHere, $a, b, c$ are in A.P.
Let $a = A - d , b = Ac = A + d$
Here, $a+b+c=18$
$A-d+A+A+d=18$
$3 A=18$
$A=6$
And,
$(a+4),(b+4),(c+36) \text { are in G.P. }$
$(6-d+4),(6+4),(6+d+36) \text { are in G.P. }$
$(10-d),(10),(42+d)$
$(10)^2=(10-d)(42+d)$
$100=420+10 d-42 d-d^2$
$d^2+32 d-320=0$
$(d+40)(d-8)=8$
$d=-40,8$
So, Numbers of $-2,6,14$, or $46,6,-34$.
View full question & answer→Question 433 Marks
Find the sum of the following geometric series:
$1,-\text{a},\text{a}^2,-\text{a}^3,\ ...\ \text{to n terms}(\text{a}\neq1)$
AnswerRewriting the sequence and sum we get,
$\text{Sum}=1-\text{a}+\text{a}^2-\text{a}^3+\text{a}^4-\text{a}^5+\ ...$
Here, $\text{r}=-\text{a and first term}=1$
$\text{Sum}=\frac{\big[1-(-\text{a}^\text{n})\big]}{1+\text{a}}$
View full question & answer→Question 443 Marks
Find the $7^{th}$ term of the G.P. is 8 times the $4^{th}$ term and $5th$ term is $48$, find the G.P.
Answer$t_7=8 t_4$
$t_5=48$
We know that $t_n=a r^{n-1}$
$a=\text { firm term }$
$r=\text { common ratio }$
$n=\text { number of terms }$
$t_7=a r^6=8\left(a r^3\right)$
$r^3=8$
$r=2$
Also,
$t_5=48$
$a r^4=48$
a $(2)^4=48$
$a=\frac{48}{16}=3$
$\therefore$ G.P. is a, $ar _{,} ar ^2, \ldots 3,6,12, \ldots$
View full question & answer→Question 453 Marks
If $a, b, c$ are in G.P., prove that:
$\text{a}^2\text{b}^2\text{c}^2\Big(\frac{1}{\text{a}^3}+\frac{1}{\text{b}^3}+\frac{1}{\text{c}^2}\Big)=\text{a}^3+\text{b}^3+\text{c}^3$
Answer$a, b, c$ are in G.P.
$a, b = ar, c = ar^2$
$\text{L.H.S}=\text{a}^2\text{b}^2\text{c}^2\Big(\frac{1}{\text{a}^3}+\frac{1}{\text{b}^3}+\frac{1}{\text{c}^3}\Big)$
$=\text{a}^2\times\text{a}^2\text{r}^2\times\text{a}^2\text{r}^4\Big(\frac{1}{\text{a}^3}+\frac{1}{\text{a}^3\text{r}^3}+\frac{1}{\text{a}^3\text{r}^6}\Big)$
$=\text{a}^6\text{r}^6\Big(\frac{\text{r}^6+\text{r}^3+1}{\text{a}^3\text{r}^6}\Big)$
$=\text{a}^3(\text{r}^6+\text{r}^3+1)$
$=\text{a}^3+\text{a}^3\text{r}^3+\text{a}^3\text{r}^6$
$=\text{a}^3+(\text{ar})^3+(\text{ar}^2)^3$
$=\text{a}^3+\text{b}^3+\text{c}^3$
$=\text{R.H.S}$
$\therefore\text{R.H.S}=\text{L.H.S}$
View full question & answer→Question 463 Marks
If a, b, c are in G.P., prove that:
$\frac{(\text{a}+\text{b}+\text{c})^2}{\text{a}^2+\text{b}^2+\text{c}^2}=\frac{\text{a}+\text{b}+\text{c}}{\text{a}-\text{b}+\text{c}}$
Answer$a, b, c$ are in G.P.$a, b = ar, c = ar^2$
$\text{L.H.S}=\frac{\text{a}+\text{b}+\text{c}}{\text{a}^2+\text{b}^2+\text{c}^2}$
$=\frac{(\text{a}+\text{ar}+\text{ar}^2)^2}{\text{a}^2+\text{a}^2\text{r}^2+\text{a}^2\text{r}^4}$
$=\frac{\text{a}^2(1+\text{r}+\text{r}^2)^2}{\text{a}^2(1+\text{r}^2+\text{r}^4)}$
$=\frac{\text{a}^2(1+\text{r}+\text{r}^2)^2}{\text{a}^2\big[\big(1+2\text{r}^2+\text{r}^4\big)-\text{r}^2\big]}$
$=\frac{\text{a}^2(1+\text{r}+\text{r}^2)^2}{\text{a}^2\big[\big(1+\text{r}^2+\text{r}\big)\big(1+\text{r}^2+\text{r}\big)\big]}$
$=\frac{\text{a}\big(1+\text{r}+\text{r}^2\big)}{\text{a}\big(1+\text{r}^2+\text{r}\big)}$
$=\frac{\text{a}+\text{ar}+\text{ar}^2}{\text{a}+\text{ar}^2-\text{ar}}$
$=\frac{\text{a}+\text{b}+\text{c}}{\text{a}-\text{b}+\text{c}}$
$=\text{R.H.S}$
$\therefore\text{R.H.S}=\text{L.H.S}$
View full question & answer→Question 473 Marks
Find the sum of the following geometric progrssions:$4,2,1,\frac{1}{2}\ ...\text{to 10 terms.}$
Answer$4, 2, 1, \frac12,\ ...10\text{ terms}$
$\text{a}=4,\text{r}=\frac{2}{4}=\frac{1}{2},\text{n}=10$
$\text{S}_\text{n}=\text{a}\frac{(1-\text{r}^{\text{n}})}{1-\text{r}}$ $[\because\text{r}<1]$
$=4\frac{1-\Big(\frac{1}{2}\Big)^{10}}{1-\frac{1}{2}}$
$=8\Big(1-\frac{1}{2^{10}}\Big)$
$=8\Big(1-\frac{1}{{1024}}\Big)$
View full question & answer→Question 483 Marks
Insert $6$ geometric means between $27$ and $\frac{1}{81}.$
Answer$6$ Geometric means between $27$ and $\frac{1}{81}.$
Let $G_1, G_2, G_3, G_4, G_5, G_6$ be $6$ geometric means between $a = 27$ and $\text{b}=\frac{1}{81}$
Then, $27, G_1, G_2, G_3, G_4, G_5, G_6$, $\frac{1}{81}$ is a G.P. with common ratio r given by
$\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{\text{n}+1}}$
$=\Bigg(\frac{\frac{1}{81}}{27}\Bigg)^{\frac{1}{6+1}}=\Big(\frac{1}{81\times27}\Big)^{\frac{1}{7}}=\Big(\frac{1}{3^7}\Big)^{\frac{1}{7}}$
$\therefore\text{G}_1=\text{ar}=27\times\Big(\frac13\Big)=9$
$\text{G}_2=\text{ar}^2=27\times\frac{1}{9}=3$
$\text{G}_3=\text{ar}^3=27\times\frac{1}{27}=1$
$\text{G}_4=\text{ar}^4=27\times\frac{1}{27\times3}=\frac13$
$\text{G}_5=\text{ar}^5=27\times\frac{1}{{3}^5}=\frac{1}{9}$
$\text{G}_6=\text{ar}^6=27\times\frac{1}{36}=\frac{1}{27}$
Hence, $9, 3, 1, \frac{1}{3},\frac19,\frac{1}{27}$ are $6$ geometric means between $27$ and $\frac{1}{81}.$
View full question & answer→Question 493 Marks
Evaluate the following:$\sum_\limits{\text{n}=1}^{11}(2+3^\text{n})$
Answer$\sum_\limits{\text{n}=1}^{11}(2+3^\text{n})$
$=(2+3^1)+(2+3^2)+(2+3^3)+\ \dots\ +(2+3^{11})$
$=2\times11+3^1+3^2+3^3+\dots+3^{11}$
$=22+\frac{3(3^{11}-1)}{(3-1)}$
$=22+\frac{3(3^{11}-1)}{2}$
$=\frac{44+3(177147-1)}{2}$
$=\frac{44+3(177146)}{2}$
$=265741$
So,
$\sum_\limits{\text{n}=1}^{11}(2+3^\text{n})=265741$
View full question & answer→Question 503 Marks
Find k such that $\text{k}+9,\text{k}-6$ and 4 from three consecutive terms of a G.P.
Answer$\text{k}+9,\text{k}-6,4 \text{ are in G.P.}$
$(\text{k}-6)^2=(\text{k}+9)4$
$\text{k}^2+36-12\text{k}=4\text{k}+36$
$\text{k}^2-16\text{k}=0$
$\text{k}(\text{k}-16)=0$
$\text{k}=0,\text{k}=16$
View full question & answer→Question 513 Marks
If $S_p$ denotes the sum of the series $1+\text{r}^{\text{p}}+\text{r}^{2\text{r}}+\ \dots\text{ to }\infty$ and $S_p$ the sum of the series $1-\text{r}^{\text{p}}+\text{r}^{\text{2p}}-\ \dots\text{ to }\infty,$ prove that $\text{S}_\text{p} + \text{S}_\text{p} = 2 \text{S}_{2\text{p}}.$
Answer$\text{S}_\text{p}=1+\text{r}^{\text{p}}+\text{r}^{2\text{p}}+\ \dots\ +\infty$
$\text{S}_\text{p}=\frac{1}{1-\text{r}^{\text{r}}}$
$\text{S}_\text{p}=1-\text{r}^\text{p}+\text{r}^{2\text{p}}+\ \dots\ +\infty$
$\text{S}_\text{p}=\frac{1}{1+\text{r}^\text{p}}$
Now,
$\text{S}_\text{p}+\text{S}_\text{p}=\frac{1}{1-\text{r}^\text{p}}+\frac{1}{1+\text{r}^{\text{p}}}$
$=\frac{2}{1-\text{r}^{2\text{p}}}$
$\text{S}_\text{p}+\text{S}_\text{p}=2\times\text{S}_{2\text{p}}$
View full question & answer→Question 523 Marks
If $a, b, c$ are in A.P. and $a, x, b$ and $b, y, c $are in G.P., show that $x2, b2, y2$ are in A.P.
Answer$a, b, c$ are in A.P.
$\Rightarrow2\text{b}=\text{a}+\text{c }\cdots(\text{i})$
$a, x, b$ are in G.P.
$\Rightarrow\text{x}^2=\text{ab }\cdots{\text{(ii}})$
$b, y, c$ are in G.P.
$\Rightarrow\text{y}^2=\text{bc}\cdots(\text{iii})$
Now, putting the value of $a$ and $c$:
$\Rightarrow2\text{b}=\frac{\text{x}^2}{\text{b}}+\frac{\text{y}^2}{\text{b}}$
$\Rightarrow2\text{b}^2=\text{x}^2+\text{y}^2$
Therefore, $x^2, b^2$ and $y^2$ are also in A.P.
View full question & answer→Question 533 Marks
The sum of terms of the G.P. 3, 6, 12, ... is 381. Find the value of n.
Answer3, 6, 12, ... n 381
$\text{a}={3},\text{r}=\frac{6}{{3}}={2},\text{n}=?,\text{S}_\text{n}=381$
We know that
$\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}$
$381=\frac{3(2)^\text{n}-1}{2-1}$
$\frac{381}{3}=2^{\text{n}}-1$
$127=2^\text{n}-1$
$128=2^\text{n}$
$2^7=2^{\text{n}}$
$\Rightarrow\text{n}=7$
View full question & answer→Question 543 Marks
Find the two numbers whose A.M. is 25 and G.M. is 20.
AnswerGiven,
A.M. = 25
G.M. = 20
Now,
$\text{A}.\text{M}.=\frac{\text{a}+\text{b}}{2}=25$
And, $\text{G}.\text{M}=\sqrt{\text{ab}}=20$
$\text{a}+\text{b}=50,\text{ab}=400$
$(\text{a}-\text{b})=\sqrt{(\text{a}+\text{b})^2-4\text{ab}}$
$=\sqrt{(50)^2-16000}$
$=\sqrt{2500-1600}$
$=\pm30$
$\text{a}-\text{b}=\pm30\\ {\text{a}+\text{b}=50}\\ \overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \ \ \text{2a} = \ \ \ 80$
$\text{a}=40$
Also, $-2\text{b}=-20$
$\text{b}=10$
$\therefore$ The numbers are 40, 10.
View full question & answer→Question 553 Marks
The ratio of the sum of first three term is to that of first $6$ terms of a G.P. is $125 : 152$. Find the common ratio.
AnswerLet Sum of first three terms $= a + ar + ar^2$
The ratio $=\frac{\text{a}+\text{ar}+\text{ar}^2}{\text{a}+\text{ar}+\text{ar}^2+\text{ar}^3+\text{ar}^4+\text{ar}^5}$
$=\frac{1+\text{r}+\text{r}^2}{1+\text{r}+\text{r}^2+\text{r}^3(1+\text{r}+\text{r}^2)}\cdots(1)$
Let $\text{A}=1+\text{r}+\text{r}^2\cdots(2)$
$\text{Ratio}=\frac{\text{A}}{\text{A}+\text{r}^3\text{A}}=\frac{125}{152}$
$\frac{1}{1+\text{r}^3}=\frac{125}{152}$
$152+125+125\text{r}^3$
$\text{r}^3=\frac{27}{125}$
$\text{r}=\frac35$
View full question & answer→Question 563 Marks
Show that the ratio of the sum of the first n terms of a G.P. to the sum of terms from $(\text{n}+1)^\text{th}\text{ to }(2\text{n})^\text{th}\text{ terms is }\frac{1}{\text{r}^\text{n}}.$
AnswerSum of first n term of G.P.
$=\text{a}+\text{a}_2+\text{a}_3+\ ...\ +\text{a}_\text{n}$
$=\text{a}+\text{ar}+\text{ar}^2+\ ...\ +\text{ar}^{\text{n}-1}$ $[\therefore\text{t}_\text{n}=\text{ar}^{\text{n}-1}]\cdots(\text{i})$
Also sum of term from
$(\text{n}+1)^{\text{th}}\text{ to }(2\text{n})^{\text{th}}\text{ term is}$
$=\text{a}_{\text{n}+1}+\text{a}_{\text{n}+1}+\dots\ +\text{a}_{2\text{n}}$
$=\text{ar}^{\text{n}}+\text{ar}^{\text{n}-1}+\ ...\ +\text{ar}^{2\text{n}-1}\cdots{(\text{i})}$
Ratio of (i) and (ii) is:
$=\frac{\text{a}+\text{ar}+\text{ar}^2+\ ...\ \text{ar}^{\text{n}-1}}{\text{ar}^{\text{n}}+\text{ar}^{\text{n}-1}+\ ...\ +\text{ar}^{2\text{n}-1}}$ $\Big[\because\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}\Big]$
$=\frac{\frac{\text{a}(1-\text{r}^{\text{n}})}{1-\text{r}}}{\frac{\text{ar}^\text{n}(1-\text{r}^\text{n})}{1-\text{r}}}$
$=\frac{1}{\text{r}^\text{n}}$
View full question & answer→Question 573 Marks
Find the sum: $\sum\limits_{\text{n}=1}^{10}\bigg\{\Big(\frac12\Big)^{\text{n}-1}+\Big(\frac15\Big)^{\text{n}+1}\bigg\}.$
Answer $\sum\limits_{\text{n}=1}^{10}\bigg\{\Big(\frac12\Big)^{\text{n}-1}+\Big(\frac15\Big)^{\text{n}+1}\bigg\}$
$=\sum\limits_{\text{n}=1}^{10}\Big(\frac12\Big)^{\text{n}-1}+\sum\limits_{\text{n}=1}^{10}\Big(\frac15\Big)^{\text{n}+1}$
$=1+\frac12+\frac{1}{2^2}+\ \cdots\ +\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\ \cdots$
$=\frac{\Big(1-\frac{1}{2^{10}}\Big)}{1-\frac12}+\frac{\frac15\Big(1-\frac{1}{5^{10}}\Big)}{1-\frac15}$
$=\frac{2^{10}-1}{2^9}+\frac{5^{10}-1}{5^{11}}$
View full question & answer→Question 583 Marks
How many terms of G.P. $3,\frac32,\frac34\cdots$ are needed to give the sum $\frac{3069}{512}?$
Answer$\text{Sum}=\frac{3069}{512}=\frac{3\big(1-\frac{1}{2^{\text{n}}}\big)}{\frac12}$
$1-\frac{1}{2^\text{n}}=\frac{3069}{512\times6}=\frac{1023}{512\times2}$
$1-\frac{1023}{1024}=\frac{1}{2^\text{n}}$
$\frac{1}{2^\text{n}}=\frac{1}{1024}$
$\text{n}=10$
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