Question 14 Marks
Let $a_n$ be the nth term of the G.P. of positive numbers. Let $\sum\limits_{\text{n}=1}^{100}\text{a}_{2\text{n}}=\alpha\text{ and}\sum\limits_{\text{n}=1}^{10}\text{a}_{2\text{n}-1}=\beta,$ such that $\alpha\neq\beta.$ Prove that the common ratio of the G.P. is $\frac{\alpha}{\beta}$
AnswerGiven: $\sum\limits_{\text{n}=1}^{100}\text{a}_{2\text{n}}=\alpha$
$\Rightarrow\text{a}_2+\text{a}_4+\text{a}_6+\ ...\ +\text{a}_{200}=\alpha\cdots(\text{i})$ Also, $\sum\limits_{\text{n}=1}^{10}\text{a}_{2\text{n}-1}=\beta,$
$\Rightarrow\text{a}_1+\text{a}_3+\text{a}_5+\ ...\ +\text{a}_{199}=\beta\cdots(\text{ii})$ Sum of G.P, $\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{n}}$
$=\text{a}=\text{a}_2,\text{r}=\text{r}^2,\text{n}=100$
$\text{ar}+\text{ar}^3+\text{ar}^5+\ ...\ +\text{ar}^{199}=\alpha$
$\text{ar}\frac{\Big(1-\big(\text{r}^2\big)^{100}\Big)}{1-\text{r}^2}=\alpha\cdots(\text{iii})$
$\text{a}+\text{ar}^2+\text{ar}^4+\ ...\ +\text{ar}^{198}=\beta$
$\frac{\text{a}\Big(1-\big(\text{r}^2\big)^{100}\Big)}{1-\text{r}^{2}}=\beta\cdots(\text{iv})$
$\text{r}(\beta)=\alpha\cdots(\text{v})$$\text{r}=\frac{\alpha}{\beta}$ [From (iv) and (v)]
View full question & answer→Question 24 Marks
If a, b, c, are in G.P., prove that: $\frac{1}{\text{a}^2+\text{b}^2},\frac{1}{\text{b}^2+\text{c}^2},\frac{1}{\text{c}^2+\text{d}^2}\text{ are in G.P.}$
Answera, b, c, d are in G.P. $\therefore\text{b}^2=\text{ac}$ $\text{ad}=\text{bc}$ $\text{c}^2=\text{bd}\cdots(1)$ $\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\Big(\frac{1}{\text{b}^2}\Big)^2+\frac{2}{\text{b}^2\text{c}^2}+\Big(\frac{1}{\text{c}^2}\Big)^2$ $\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\Big(\frac{1}{\text{ac}}\Big)^2+\frac{1}{\text{b}^2\text{c}^2}+\frac{1}{\text{b}^2\text{c}^2}+\Big(\frac{1}{\text{bd}}\Big)^2$ [Using (1)] $\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\frac{1}{\text{a}^2\text{c}^2}+\frac{1}{\text{a}^2\text{d}^2}+\frac{1}{\text{b}^2\text{c}^2}+\frac{1}{\text{b}^2\text{d}^2}$ [Using (1)]$\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\frac{1}{\text{a}^2}\Big(\frac{1}{\text{c}^2}+\frac{1}{\text{d}^2}\Big)+\frac{1}{\text{b}^2}\Big(\frac{1}{\text{c}^2}+\frac{1}{\text{d}^2}\Big)$
$\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\Big(\frac{1}{\text{a}^2+\text{b}^2}\Big)\Big(\frac{1}{\text{c}^2}+\frac{1}{\text{d}^2}\Big)$
$\therefore\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big),\Big(\frac{1}{\text{c}^2+\text{d}^2}\Big)\text{ and are also in G.P.}$
View full question & answer→Question 34 Marks
The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is $87\frac{1}{2}.$ Find them.
AnswerLet the three number in G.P. be $\frac{\text{a}}{\text{r}},\text{a},\text{ar}$ then product of these numbers $\Big(\frac{\text{a}}{\text{r}}\Big)(\text{a})(\text{ar})$ $\Rightarrow\text{a}^3=125=5^3$ $\text{a}=5$ Also, sum of these product in pair $\Big(\frac{\text{a}}{\text{r}}\Big)(\text{a})+(\text{a})(\text{ar})+\Big(\frac{\text{a}}{\text{r}}\Big)(\text{ar})$ $=87\frac{1}{2}=\frac{195}{2}$ $=(5)^2\Big(\frac{1+\text{r}^2+\text{r}}{\text{r}}\Big)=\frac{195}{2}$ $1+\text{r}^2+\text{r}=\Big(\frac{195}{2\times25}\Big)^\text{r}$ $2(1+\text{r}^2+\text{r})=\frac{39}{5}\text{r}$ $10+10\text{r}^2+10\text{r}=39\text{r}$ $10\text{r}^2-25\text{r}-4\text{r}+10=0$ $5\text{r}(2\text{r}-5)-2(2\text{r}-5)=0$ $\text{r}=\frac{5}{2},\frac{2}{5}$ $\therefore\text{G.P. is }\frac{\text{a}}{\text{r}},\text{a},\text{ar}$ $10, 5,\frac{5}{2}, \ ... \text{or}\frac{5}{2},5,10\ ...$
View full question & answer→Question 44 Marks
If a, b, c are in G.P., prove that: $\frac{\text{a}^2+\text{ab}+\text{b}^2}{\text{bc}+\text{ca}+\text{ab}}=\frac{\text{b}+\text{a}}{\text{c}+\text{b}}$
Answer$a, b, c$ are in G.P. $a, b=a r, c=a r^2$
$\frac{\text{a}^2+\text{ab}+\text{b}^2}{\text{bc}+\text{ca}+\text{ab}}=\frac{\text{b}+\text{a}}{\text{c}+\text{b}}$
$\frac{\text{a}^2+\text{a}(\text{ar})+\text{a}^2\text{r}^2}{(\text{ar})\big(\text{ar}^2\big)+\big(\text{ar}^2\big)\text{a}+\text{a}(\text{ar})}=\frac{\text{ar}+\text{a}}{\text{ar}^2+\text{ar}}$
$\frac{\text{a}^2\big(1+\text{r}+\text{r}^2\big)}{\text{a}^2(\text{r}^3+\text{r}^2+\text{r})}=\frac{1+\text{r}}{\text{r}(1+\text{r})}$ $\frac{1}{\text{r}}=\frac{1}{\text{r}}$
$\text{L.H.S}=\text{R.H.S}$ So, $\frac{\text{a}^2+\text{ab}+\text{b}^2}{\text{bc}+\text{ca}+\text{ab}}=\frac{\text{b}+\text{a}}{\text{c}+\text{b}}$
View full question & answer→Question 54 Marks
If a and b are the roots of $\text{x}^2-3\text{x}+\text{p}=0$ and c, d are roots $\text{x}^2-12\text{x}+\text{q}=0,$ where a, b, c, d from a G.P. Prove that (q + p) : (q - p) = 17 : 15.
AnswerGiven, a, b are roots of the equation $\text{x}^2-3\text{x}+\text{p}=0$ $\Rightarrow\text{a}+\text{b}=3,\text{ab}=\text{p}$ And c, d are roots of the equation $\text{x}^2-12\text{x}+\text{q}=0$ $\Rightarrow\text{c}+\text{d}=12, \text{cd}=\text{q}$ Let b = ar, c = ar2 and d = ar3, then a + b = 3 and c + d = 12 $\text{a}(1+\text{r})=3\text{ and ar}^2(1+\text{r})=12$ $\Rightarrow\frac{\text{ar}^2(1+\text{r})}{\text{a}(1+\text{r})}=\frac{12}{3}$ $\Rightarrow\text{r}=2$ And $\text{a}(\text{r}+2)=3$ $\Rightarrow\text{a}=1$ $\text{p}=\text{ab}$ $\text{p}=\text{a}\times\text{ar}$ $\text{p}=2$ $\text{q}=\text{cd}$ $=\text{ar}^2\times\text{ar}^3$ $\text{a}=32$ $\frac{\text{q}+\text{p}}{\text{q}-\text{p}}=\frac{32+2}{32-2}$ $=\frac{34}{30}$ $(\text{q}+\text{p}):(\text{q}-\text{p})=17:15$
View full question & answer→Question 64 Marks
If a, b, c, d are in G.P., prove that: $(\text{b}+\text{c})(\text{b}+\text{d})=(\text{c}+\text{a})(\text{c}+\text{d})$
Answera, b and c are in G.P.$\therefore\text{b}^2=\text{ac }\cdots(1)$
$\text{L.H.S}=({\text{b}+\text{a})(\text{b}+\text{d})}$ $=\text{b}^2+\text{bd}+\text{bc}+\text{cd}$ $=\text{ac}+\text{c}^2+\text{ad}+\text{cd}$ $[\text{Using (1)}]$ $=\text{c}(\text{a}+\text{c})+\text{d}(\text{a}+\text{c})$ $=(\text{c}+\text{a})(\text{c}+\text{d})$ $=\text{R.H.S}$$\therefore\text{R.H.S}=\text{L.H.S}$
View full question & answer→Question 74 Marks
Find the $4^{th}$ term from the end of the G.P. $\frac12,\frac16,\frac{1}{18}\frac{1}{54},\ \dots,\ \frac{1}{4374}.$
Answer$\frac12,\frac16,\frac{1}{18}\frac{1}{54},\ \dots,\ \frac{1}{4374}.$ $\text{a}=\frac{1}{2},\text{l}=\frac{1}{4374},\text{r}=\frac{\text{t}_{\text{n}-1}}{\text{t}_\text{n}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac16}{\frac12}=\frac13$ Term from the end is: $\text{a}_\text{n}=\text{l}\Big(\frac{1}{\text{r}}\Big)^{\text{n}-1}$ $\text{t}_4=\Big(\frac{1}{4374}\Big)(3)^{\text{n}-1}$ $=\frac{1}{4374}\times3^3$ $=\frac{1}{162}$ $\therefore 4^{th}$ term from the end is $=\frac{1}{162}$
View full question & answer→Question 84 Marks
If $p^{th}, q^{th}$ and $r^{th}$ terms of an A.P. and G.P. are both a, b and c respectively, show that $a^{b-c} b^{c-a} c^{a-b} = 1$.
AnswerLet the A.P. be A, A + D, A + 2D, ... and G.P. be $x, xR, xR^2$, ...then $a = A + (p - 1)D, b = A + (q - 1)D, c = A + (r - 1)D \Rightarrow a - b = (p - q)D, b - c = (q - r)D, c - a - (r - p)D$
Also $\text{a} = \text{xR}^{\text{p}-1}, \text{b} = \text{xR}^{\text{q}-1}, \text{c} = \text{xR}^{\text{r}-1}$ Hence $\text{a}^{\text{b}-\text{c}}.\text{b}^{\text{c}-\text{a}}.\text{c}^{\text{a}-\text{b}} = \big(\text{xR}^{\text{p}-1}\big)^{(\text{q}-\text{r})}\text{D}.\big(\text{xR}^{\text{q}-1}\big)\text{D}.\big(\text{xR}^{\text{r}-1}\big)\big(\text{p}-\text{q}\big)\text{D}$
$= \text{x}^{(\text{q}-\text{r}+\text{r}-\text{p}+\text{p}-\text{q})\text{D}}.\text{R}^{\big[(\text{p}-1)(\text{q}-\text{r})+(\text{q}-1)(\text{r}-\text{1})+(\text{r}-1)(\text{p}-\text{q})\big]\text{D}}$$=\text{x}0. \text{R}0 = 1.1 = 1$
View full question & answer→Question 94 Marks
Find three numbers in G.P. whose sum is 38 and their product is 1728.
AnswerLet the three number be $a, ar, ar^2$ in G.P., where a is first teror and r is the common ratio.Then,
$\text{a} + \text{ar} +\text{ar}^2 = 38$
$\text{a} (1 + \text{r} + \text{r}^2) = 38\cdots(\text{i})$
and
$\text{(a)}\text{(ar)}\text{(ar)}^2=1728$
$\text{a}^3\text{r}^3=1728=4^33^3=(12)^3$
$\text{a}^3=\frac{12^3}{\text{r}^3}\Rightarrow\frac{12}{\text{r}}=\text{a}$
Putting $\text{a}=\frac{12}{\text{r}}\text{ in }(\text{i})$
$\frac{12}{\text{r}}(1 + \text{r} + \text{r}^2)=38$
$12+12\text{r}+12\text{r}^2=38\text{r}$
$12\text{r}^2-26\text{r}+12=0$
$6\text{r}^2-13\text{r}+6=0$
$6\text{r}^2-9\text{r}-4\text{r}+6=0$
$3\text{r}(3\text{r}-3)-2(3\text{r}-3)=0$
$\text{r}=\frac{3}{2},\frac{2}{3}$
$\text{a}=\frac{12}{\frac{3}{2}}=8\text{ or }\frac{12}{\frac23}=18$
$\therefore$ G.P. is 8, 12, 18.
View full question & answer→Question 104 Marks
Find sum of these numbers in G.P. is $21$ and the sum of their aquares is $189$. Find the numbers.
AnswerLet the required numbers be $a, ar,$ and $ar^2$. Sum of the numbers = 21 $\Rightarrow \text{a} + \text{ar} + \text{ar}^2=21$
$\Rightarrow \text{a} (1+ \text{r} + \text{r}^2)=21\dots(\text{i})$ Sum of the squares of the numbers = 189 $\Rightarrow\text{a}^2+(\text{a}\text{r})^2+(\text{a}\text{r})^2=189$
$\Rightarrow\text{a}^2+(1+\text{r}^2+\text{r}^4)=189\dots(\text{ii})$ Now, $\text{a}(1+\text{r}+\text{r}^2)=21$ [From (i)] Squaring both the sides $\Rightarrow\text{a}^2(1+\text{r}+\text{r}^2)=441$
$\Rightarrow\text{a}^2(1+\text{r}^2+\text{r}^4)+2\text{a}^2\text{r}(1+\text{r}+\text{r}^2)=441$
$\Rightarrow189+2\text{ar}\{\text{a}(1+\text{r}+\text{r}^2)\}=441$ [Using (ii)] $\Rightarrow189+2\text{ar}\times21=441$ [Using (i)] $\Rightarrow\text{ar}=6$
$\Rightarrow\text{a}=\frac{6}{\text{r}}\cdots(\text{iii})$ Putting $\text{a}=\frac{6}{\text{r}}\text{ in (i)}$
$\frac{6}{\text{r}}(1+\text{r}+\text{r}^2)=21$
$\Rightarrow\frac{6}{\text{r}}+6+6\text{r}=21$
$\Rightarrow6\text{r}^2+6\text{r}+6=21\text{r}$
$\Rightarrow6\text{r}^2-15\text{r}+6=0$
$\Rightarrow3(2\text{r}^2-5\text{r}+2)=0$
$\Rightarrow2\text{r}^2-5\text{r}+2=0$
$\Rightarrow(2\text{r}-1)(\text{r}-2)=0$
$\Rightarrow\text{r}=\frac12,2$ Putting $\text{r}=\frac12$ in $\text{a}=\frac{6}{\text{r}},$ we get a = 12. So, the numbers are 12, 6 and 3. Putting r = 2 in $\text{a}=\frac{6}{\text{r}},$ we get a = 3. So, the Numbers are 3, 6 and 12. Hence, the numbers that are in G.P. are 3, 6 and 12.
View full question & answer→Question 114 Marks
Find the $4^{th}$ term of the G.P. is $27$ and the 7th term is $729$, find the G.P.
Answer$\mathrm{t}_4=27 \mathrm{~t}_7=729$ We know that $\mathrm{t}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1} \mathrm{t}_4=a \mathrm{r}^3=27 \mathrm{t}_7=a \mathrm{r}^6=729$
Now, $\frac{\mathrm{t}_7}{\mathrm{t}_4}=\frac{\mathrm{ar} \mathrm{r}^6}{\mathrm{ar} \mathrm{r}^3}=\mathrm{r}^3=\frac{729}{27} \mathrm{r}^3=\left(\frac{9}{3}\right)^3$
$r^3=3^3 r=3 t_4=a r^3=27 a(3)^3=27 a(27)=27 \mathrm{a}=1$
Now G.P. is $a, a r, a r^2, \ldots 1,3,9, \ldots$
View full question & answer→Question 124 Marks
If the $4^{th}$, $10^{th}$ and $16^{th}$ terms of a G.P. are x, y and z respectiveiy. Prove that x, y, z are in G.P.
Answer$\text{a}_4=\text{x}$ $\Rightarrow\text{ar}^3=\text{x}$ Also, $\text{a}_{16}=\text{y}$ $\Rightarrow\text{ar}^9=\text{y}$ And, $\text{a}_{16}=\text{z}$ $\Rightarrow\text{ar}^{15}=\text{z}$ $\because\frac{\text{z}}{\text{x}}=\frac{\text{ar}^9}{\text{ar}^3}=\text{r}^6$ And, $\frac{\text{z}}{\text{y}}=\frac{\text{ar}^{15}}{\text{ar}^9}=\text{r}^6$ $\therefore\frac{\text{y}}{\text{x}}=\frac{\text{z}}{\text{y}}$ $\therefore\text{x},\text{y and z are in G.P.}$
View full question & answer→Question 134 Marks
If the A.M. of two positive numbers a and b (a > b) is twice their geometric mean. Prove that: $\text{a} : \text{b} = \big(2 +\sqrt{3}\big):\big(2-\sqrt{3}\big).$
AnswerAM = 2GM $\therefore\frac{\text{a}+\text{b}}{2}=2\sqrt{\text{ab}}$ $\Rightarrow\text{a}+\text{b}=4\sqrt{\text{ab}}$ Squaring both the sides: $\Rightarrow(\text{a}+\text{b})^2=\big(4\sqrt{\text{ab}}\big)^2$ $\Rightarrow\text{a}^2+2\text{ab}+\text{b}^2=16\text{ab}$ $\Rightarrow\text{a}^2-14\text{ab}+\text{b}^2=0$ Using the quadratic formula: $\Rightarrow\text{a}= {-(-14\text{b}) \pm \sqrt{(-14\text{b})^2-4\times1\times\text{b}^2} \over 2\times1}$ $[\because$ a is positive number$]$ $\Rightarrow\text{a}= {14\text{b} +2\text{b} \sqrt{49-1} \over 2}$ $\Rightarrow\text{a}=\text{b}\big(7+4\sqrt{3}\big)$ $\Rightarrow\frac{\text{a}}{\text{b}}=7+4\sqrt{3}$ $\Rightarrow\frac{\text{a}}{\text{b}}=4+3+2\times2\times\sqrt{3}$ $\Rightarrow\frac{\text{a}}{\text{b}}=\big(2+\sqrt{3}\big)^2$ $\Rightarrow\frac{\text{a}}{\text{b}}=\frac{\big(2+\sqrt{3}^2\big(2-\sqrt{3}\big)}{\big(2-\sqrt{3}\big)}$ $\Rightarrow\frac{\text{a}}{\text{b}}\frac{\big(2+\sqrt{3}\big)(4-3)}{\big(2-\sqrt{3}\big)}$ $\therefore\frac{\text{a}}{\text{b}}=\frac{\big(2+\sqrt{3}\big)}{\big(2-\sqrt{3}\big)}$
View full question & answer→Question 144 Marks
If a, b, c, are in G.P., prove that: $(\text{a}^2+\text{b}^2),(\text{b}^2+\text{c}^2),(\text{c}^2+\text{d}^2)\text{ are in G.P.}$
Answera, b, c, d are in G.P. $\therefore\text{b}^2=\text{ac}$ $\text{ab}=\text{bc}$ $\text{c}^2=\text{bd}\cdots(1)$ Now, $(\text{b}^2+\text{c}^2)^2=\big(\text{b}^2\big)^22\text{b}^2\text{c}^2+\big(\text{c}^2\big)^2$ $\Rightarrow\big(\text{b}^2+\text{c}^2\big)^2=\big(\text{ac}\big)^2+\text{b}^2\text{c}^2+\text{b}^2\text{c}^2+\big(\text{bd}\big)^2$ [Using (1)] $\Rightarrow\big(\text{b}^2+\text{c}^2\big)^2=\text{a}^2\text{c}^2+\text{a}^2\text{d}^2+\text{b}^2\text{c}^2+\text{b}^2\text{d}^2$ [Using (1)] $\Rightarrow\big(\text{b}^2+\text{c}^2\big)^2=\text{a}^2\big(\text{c}^2+\text{d}^2\big)+\text{b}^2\big(\text{c}^2+\text{d}\big)^2$ $\Rightarrow\big(\text{b}^2+\text{c}^2\big)^2=\big(\text{a}^2+\text{b}^2\big)\big(\text{c}^2+\text{d}^2\big)$ $\therefore\big(\text{a}^2+\text{b}^2\big),\big(\text{c}^2+\text{d}^2\big)\text{ and }\big(\text{b}^2+\text{c}^2\big)\text{ are also in G.P.}$
View full question & answer→Question 154 Marks
If one A.M., A and two geometric means G1 and G2 inserted between any two positive number, show that $\frac{\text{G}^2_1}{\text{G}_2}+\frac{\text{G}^2_2}{\text{G}_1}=2\text{A}.$
AnswerLet the two positive numbers be a and b. a, A and b are in A.P. $\therefore2\text{A}=\text{a}+\text{b }\cdots(\text{i})$ Also, a, $G_1, G_2$ and b are in G.P. $\therefore\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{3}}$ Also, $G_1 = ar$ and $G_2 = ar^2$. ...(ii) Now, $\text{L.H.S}=\frac{\text{G}^2_1}{\text{G}_2}+\frac{\text{G}^2_2}{\text{G}_1}$ $=\frac{(\text{ar})^2}{\text{ar}^2}+\frac{\big(\text{ar}^2\big)^2}{\text{ar}}$ [Using (ii)] $=\text{a}+\text{a}\Bigg(\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{3}}\Bigg)^3$ $=\text{a}+\text{a}\Big(\frac{\text{b}}{\text{a}}\Big)$ $=\text{a}+\text{b}$ $=2\text{A}$ $=\text{R.H.S}$ [Using (i)]
View full question & answer→Question 164 Marks
A G.P. consits of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying the odd places. Find the common ratio the G.P.
AnswerLet the G.P. be 2n, 2, 2n + 4, ...Then, $\text{S}_\text{n}=\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1},\text{a}=2\text{n},\text{r}=2$
$\therefore\text{S}_\text{n}=\frac{2\text{n}(2^\text{n}-1)}{2-1}=2\text{n}^{\text{n}+1}-2\text{n}$
Then the G.P. of odd term
$\text{a}_1+\text{a}_3+\text{a}_5+\ ...\ \text{a}_{2\text{n}-1}$
Accourding to the question
Sum of all terms = 5(sum of terms occupying the odd places)
$\text{a}_1+\text{a}_2+\text{a}_3+\ ...\ +\text{a}_{2\text{n}}=5(\text{a}_1+\text{a}_3+\text{a}_5+\ ...\ +\text{a}_{2\text{n}-1})$
$\text{a}+\text{ar}+\text{ar}^2+\ ...\ +\text{ar}^{2\text{n}-1}=5\big(\text{a}+\text{ar}^2+\text{ar}^4+\ ...\ +\text{ar}^{2\text{n}-1}\big)$
$\frac{\text{a}(1-\text{r}^{2\text{n}})}{1-\text{r}}=5\Bigg(\frac{\text{a}\big(1-(\text{r})^2\big)^{\text{n}}}{1-\text{r}^2}\Bigg)$
$\frac{\text{a}}{1-\text{r}}$ is cancelled on both side
$1-\text{r}^{2\text{n}}=\frac{5\big(1-\text{r}^{2\text{n}}\big)}{1+\text{r}}$
$1+\text{r}-\text{r}^{2\text{n}}-\text{r}^{2\text{n}+1}=5-5\text{r}^{2\text{n}}$
$\text{r}^{2\text{n}-1}-4\text{r}^{2\text{n}}-\text{r}+4=0$
$\text{r}^{2\text{n}}(\text{r}-4)-1(\text{r}-4)=0$
$\text{r}^{2\text{n}}=1,\text{r}=4$
$\Rightarrow\text{r}=4$
View full question & answer→Question 174 Marks
The sum of first two terms of an infinite G.P. is 5 and each term is three times the sum of the succeeding terms. Find the G.P.
Answer$\text{a}+\text{ar}=5$ $\text{a}(1+\text{r})=5\cdots(1)$ $\text{a}_\text{n}=3\big(\text{a}_{\text{n}+1}+\text{a}_{\text{n}+2}+\text{a}_{\text{n}+3}+\ \dots\big)$ $\text{ar}^{\text{n}-1}=3\big(\text{ar}^\text{n}+\text{ar}^{\text{n}+1}+\text{ar}^{\text{n}+2}+\dots\big)$ $\text{ar}^{\text{n}-1}=3\text{ar}^{\text{n}}\big(1+\text{r}+\text{r}^2+\dots\infty)$ $1=3\text{r}\Big(\frac{1}{1-\text{r}}\Big)$ $1-\text{r}=3\text{r}$ $1=4\text{r}$ $\text{r}=\frac14$ $\text{a}(1+\text{r})=5$ $\text{a}\Big(\frac{5}{4}\Big)=5$ $\text{a}=4$ $\text{G.P. is }1,\frac{1}{4},\frac{1}{16},\ \dots$
View full question & answer→Question 184 Marks
The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is 1. Find the common ratio and the terms.
AnswerLet the terms of the G.P. be $\frac{\text{a}}{\text{r}},\text{a}$ and ar. $\therefore$ Product of the G.P. = 1 $\Rightarrow\text{a}^3=1$ $\Rightarrow\text{a}=1$ Now, sum of the G.P. $=\frac{39}{10}$ $\Rightarrow\frac{\text{a}}{\text{r}}+\text{a}+\text{ar}=\frac{39}{10}$ $\Rightarrow\text{a}\Big(\frac{1}{\text{r}}+1+\text{r}\Big)=\frac{39}{10}$ $\Rightarrow1\Big(\frac{1}{\text{r}}+1+\text{r}\Big)=\frac{39}{10}$ $\Rightarrow10\text{r}^2-10\text{r}+10=39\text{r}$ $\Rightarrow10\text{r}^2-29\text{r}+10=0$ $\Rightarrow10\text{r}^2-25\text{r}-4\text{r}+10=0$ $\Rightarrow5\text{r}(2\text{r}-5)-2(2\text{r}-5)=0$ $\Rightarrow(5\text{r}-2)(2\text{r}-5)=0$ $\Rightarrow\text{r}=\frac{2}{5},\frac{5}{2}$ Hence, Putting the values of a and r, the required numbers are $\frac{5}{2},1,\frac25\text{ or }\frac{2}{5},1 \text{ and }\frac{5}{2}.$
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If a, b, c, d are in G.P., prove that: $\big(\text{a}^2+\text{b}^2+\text{c}^2\big),\big(\text{ab}+\text{bc}+\text{cd}\big),\big(\text{b}^2+\text{c}^2+\text{d}^2\big)\text{ are in G.P.}$
Answera, b, c, d are in G.P. $\therefore\text{b}^2=\text{ac}$ $\text{ad}=\text{bc}$ $\text{c}^2=\text{bd}\cdots(1)$ $(\text{ab}+\text{bc}+\text{cd})^2=\big(\text{ab}\big)^2+\big(\text{bc}\big)^2+\big(\text{cd}\big)^2+2\text{ab}^2\text{c}+2\text{bc}^2\text{d}+2\text{abcd}$ $(\text{ab}+\text{bc}+\text{cd})^2=\text{a}^2\text{b}^2+\text{b}^2\text{c}^2+\text{c}^2\text{d}^2+\text{ab}^2\text{c}+\text{bc}^2\text{d}+\text{abcd}+\text{abcd}$ $\Rightarrow\big(\text{ab}+\text{bc}+\text{cd}\big)^2=\text{a}^2\text{b}^2+\text{b}^2\text{c}^2+\text{c}^2\text{d}^2+\text{b}^2\big(\text{b}^2\big)\\+\text{ac}(\text{ac})+\text{c}^2(\text{c})^2+\text{bd}(\text{bd})+\text{bc}(\text{bc})+\text{ad}(\text{ad})$ $$$\Rightarrow(\text{ab}+\text{bc}+\text{cd})^2=\text{a}^2\text{b}^2+\text{b}^2\text{c}^2+\text{a}^2\text{d}^2+\text{b}^4+\text{b}^2\text{c}^2+\text{b}^2\text{d}^2+\text{c}^2\text{b}^2+\text{c}^4+\text{c}^2\text{d}^2$ $\Rightarrow(\text{ab}+\text{bc}+\text{cd})^2=\text{a}^2\big(\text{b}^2+\text{c}^2+\text{d}^2\big)+\text{b}^2\big(\text{b}^2+\text{c}^2+\text{d}^2\big)+\text{c}^2\big(\text{b}^2+\text{c}^2+\text{d}^2\big)$ $\Rightarrow(\text{ab}+\text{bc}+\text{cd})^2=\big(\text{b}^2+\text{c}^2+\text{d}^2\big)\big(\text{a}^2+\text{b}^2+\text{c}^2\big)$ $\therefore\big(\text{a}^2+\text{b}^2+\text{c}^2\big),(\text{ab}+\text{bc}+\text{cd})\text{ and }\big(\text{b}^2+\text{c}^2+\text{d}^2\big)\text{ are also in G.P.}$
View full question & answer→Question 204 Marks
The fifth term of a G.P. is 81 whereas its secound term is 24. Find the series and sum of its first eight terms.
AnswerFifth term of series is: $\text{ar}^{5-1}=81\cdots(1)$ Second term of series is: $\text{ar}=24\cdots(2)$ Deviding (2) by (1) we get, $\frac{\text{ar}}{\text{ar}^4}=\frac{24}{81}=\frac{8}{27}$ $\text{r}^3=\frac{27}{8}$ $\text{r}=\frac32$ Substituting r in (2), we get, $\text{a}=\frac{24\times2}{3}=16$ $\text{Sum}=\frac{16\Big[\big(\frac32\big)^8-1\Big]}{\frac32-1}$ $=\frac{16\big[3^8-2^8\big]}{2^7}$ $=\frac{6305}{8}$
View full question & answer→Question 214 Marks
The $4^{th}$ and $7^{th}$ terms of G.P. are $\frac{1}{27}\text{ and }\frac{1}{729}$ respectively. Find the sum of n terms of the G.P.
Answer$\text{t}_4=\frac{1}{27},\text{t}_7=\frac{1}{729},\text{t}_\text{n}=\text{ar}^{\text{n}-1}$ Where tn = nth term, r = common difference, n = number of terms. $\text{t}_4=\text{ar}^3=\frac{1}{27}\cdots{\text{(i})}$ $\text{t}_7=\text{ar}^6=\frac{1}{729}\cdots{\text{(ii})}$ Dividing (ii) by (i), we get $\frac{\text{t}_7}{\text{t}_4}=\frac{\text{ar}^6}{\text{ar}^3}=\text{r}^3=\frac{27}{729}=\frac{1}{27},\text{r}=\frac13$ Sum of n term $=\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}\cdots{(\text{i})}$ When, $\text{r}=3, \text{t}_4=\text{ar}^3=\frac{1}{27}$ $\text{a}\Big(\frac13\Big)^3=\frac{1}{27}$ $\text{a}=1$ Substituting $\text{a}=1,\text{r}=\frac13\text{ in (i)}$ $\text{S}_\text{n}=\frac{1\Big(1-\big(\frac{1}{3}\big)^\text{n}\Big)}{1-\frac13}$ $=\frac{1-\big(\frac13\big)^\text{n}}{\frac23}$ $=\frac32\Big(1-\Big(\frac13\Big)^\text{n}\Big)$
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If $a, b, c$, are in G.P., prove that the following are also in G.P. $\text{a}^2+\text{b}^2,\text{ab}+\text{bc},\text{b}^2+\text{c}^2$
Answera, b, c are in G.P. $a, b = ar, c = ar^2 (\text{ab}+\text{bc})^2=\big(\text{a}^2+\text{b}^2\big)\big(\text{b}^2+\text{c}^2\big)$
$\Big(\text{a}\times\text{ar}+\text{ar}\times\text{ar}^2\big)^2\big(\text{a}^2+(\text{ar})^2\big)\big((\text{ar})^2+\big(\text{ar}^2\big)^2\Big)$
$\big(\text{a}^2\text{r}+\text{a}^2\text{r}^3\big)^2=\big(\text{a}^2+\text{a}^2\text{r}^2\big)\big(\text{a}^2\text{r}^2+\text{a}^2\text{r}^4\big)$
$\text{a}^4\big(\text{r}+\text{r}^3\big)^2=\text{a}^4\big(1+\text{r}^2\big)\big(\text{r}^2+\text{r}^4\big)$
$\text{a}^4\big(\text{r}+\text{r}^2\big)^2=\text{a}^4\big(1+\text{r}^2\big)\text{r}^2\big(1+\text{r}^2\big)$
$\text{a}^4\text{r}^2\big(1+\text{r}^2\big)^2=\text{a}^4\big(1+\text{r}^2\big)\text{r}^2\big(1+\text{r}^2\big)$
$\text{a}^4\text{r}^2\big(1+\text{r}^2\big)^2=\text{a}^4\text{r}^2\big(1+\text{r}^2\big)$
$\text{L.H.S}=\text{R.H.S}$
$\Rightarrow(\text{ab}+\text{bc})^2=\big(\text{a}^2+\text{b}^2\big)\big(\text{b}^2+\text{c}^2\big)$
$\Rightarrow\big(\text{a}^2+\text{b}^2\big),(\text{ab}+\text{bc}),\big(\text{b}^2+\text{c}^2\big)\text{ are in G.P.}$
View full question & answer→Question 234 Marks
Find the sum of the following geometric series:$(\text{x}+\text{y})+(\text{x}^2+\text{xy}+\text{y}^2)+(\text{x}^3+\text{x}^2\text{y}+\text{xy}^2+\text{y})+\ ...\text{ to n terms;}$
Answer$(\text{x}+\text{y})+(\text{x}^2+\text{xy}+\text{y}^2)+(\text{x}^3+\text{x}^2\text{y}+\text{xy}^2+\text{y})+\ ...\text{ to n terms;}$ $\text{S}_\text{n}=\text{x}(\text{x}+\text{y})+\text{x}^2(\text{x}^2+\text{y}^2)+\ ...\ +\text{x}^\text{n}(\text{x}^\text{n}+\text{y}^\text{n})$ $=(\text{x}^2+\text{xy})+(\text{x}^4+\text{x}^2\text{y}^2)+\ ...\ +(\text{x}^{2\text{n}}+\text{x}^\text{n}\text{y}^\text{n})$ $=(\text{x}^2+\text{x}^4+\ ...\ +\text{x}^{2\text{n}})+(\text{xy}+\text{x}^2\text{y}^2+\ ...\ +\text{x}^{\text{n}}\text{y}^\text{n})\cdots(1)$ First term of (1) is a G.P. with $\text{A}=\text{x}^2,\text{R}=\text{x}^2,\text{N}=\text{n}$ $\Rightarrow\text{S}_\text{n1}=\frac{\text{x}^2(1-\text{x}^{2\text{n}})}{1-\text{x}^2}\cdots(2)$ Second term of (1) is a G.P. with $\text{A}=\text{xy},\text{R}=\text{xy},\text{N}=\text{n}$ $\Rightarrow\text{S}_\text{n2}=\frac{\text{xy}(1-(\text{xy}^\text{n}))}{1-\text{xy}}\cdots(3)$ $\therefore\text{S}_\text{n}=\text{S}_\text{n1}+\text{S}_\text{n2}=\frac{\text{x}^2(1-\text{x}^{2\text{n}})}{1-\text{x}^2}+\frac{\text{xy}(1-(\text{xy})^\text{n})}{1-\text{xy}}$ $\Rightarrow\text{S}_\text{n}=\frac{\text{x}^2(1-\text{x}^{2\text{n}})}{1-\text{x}^2}+\frac{\text{xy}(1-\text{x}^\text{n}\text{y}^\text{n})}{1-\text{xy}}$
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Find three numbers in G.P. whose sum is $65$ and whose product is $3375$.
AnswerLet the three number in G.P. be $\frac{\text{a}}{\text{r}},\text{a},\text{ar}$ Sum of these numbers = $\frac{\text{a}}{\text{r}}+\text{a}+\text{ar}=65$ 3375 = Product of these numbers $3375=\Big(\frac{\text{a}}{\text{r}}\Big)(\text{a})(\text{ar})=\text{a}^3$
$\text{a}^3=(5)^3\times(3)^3=(15)^3$
$\Rightarrow\text{a}=15$
$\text{a}\Big(\frac{1}{\text{r}}+1+\text{r}\Big)=65$
$15\Big(\frac{1}{\text{r}}+1+\text{r}\Big)=\frac{65}{15}=\frac{13}{3}$
$3+3\text{r}+3\text{r}^2=13\text{r}$
$3\text{r}^2-10\text{r}+3=0$
$3\text{r}^2-\text{r}-9\text{r}+3=0$
$\text{r}(3\text{r}-1)-3(3\text{r}-1)=0$
$\text{r}=3,\frac{1}{3}\ \text{r}=\frac{1}{3}\text{ or}\text{ r}=3$
$\therefore$ G.P. is $a, ar, ar^2 \therefore$ G.P. is $45, 15, 5$ or $5, 15, 45$
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The $4^{th}$ term of a G.P. is square of its second term, and the first firm is -3. Find its $7^{th}$ term.
AnswerLet r be the common ratio of the given G.P. Than, $\text{a}_4=(a_2)^2$ [Given] Now, $\text{ar}^3=\text{a}^2\text{r}^2$ $\Rightarrow\text{r}=\text{a}$. $\Rightarrow\text{r}=-3$ [Putting a = -3] $\therefore\text{a}_7=\text{ar}^6$ $\Rightarrow\text{a}_7=(-3)(-3)^6$ [Putting a = -3 and r = -3] $\Rightarrow\text{a}_7=(-3)(-729)$ $\Rightarrow\text{a}_7=-2187$ Thus, the 7th term of the G.P. is -2187.
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If a, b, c, are in G.P., prove that the following are also in G.P. $\text{a}^3,\text{b}^3,\text{c}^3$
Answera, b, c are in G.P. $a, b = ar, c = ar^2 \big(\text{b}^3\big)^2=\text{a}^3\text{c}^3$
$\big((\text{ar})^3\big)^2=\text{a}^3\big(\text{ar}^2\big)^3$
$\text{a}^6\text{r}^6=\text{a}^3\big(\text{a}^3\text{r}^6\big)$
$\text{a}^6\text{r}^6=\text{a}^6\text{r}^6$
$\text{L.H.S}=\text{R.H.S}$
$\Rightarrow\big(\text{b}^3\big)^2=\text{a}^3\text{c}^3$ So, $\Rightarrow\text{a}^3,\text{b}^3,\text{c}^3\text{ are in G.P.}$
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If a, b, c, d are in G.P., prove that: $\frac{\text{ab}-\text{cd}}{\text{b}^2-\text{c}^2}=\frac{\text{a}+\text{c}}{\text{b}}$
Answera, b, c are in G.P.$\therefore\text{b}^2=\text{ac }\cdots{1}$
$\text{L.H.S}={\text{a}\big(\text{b}^2+\text{c}^2\big)}$
$=\text{ab}^2+\text{ac}^2$
$=\text{a}(\text{ac})+\text{c}\big(\text{b}^2\big)$ $[\text{Using (1)}]$
$=\text{c}(\text{a}^2+\text{b}^2)$
$=\text{R.H.S}$
$\therefore\text{R.H.S}=\text{L.H.S}$
View full question & answer→Question 284 Marks
If a, b, c are three distinct real nimbers in G.P. and a + b + c = xb, then proved that either x < -1 or x > 3.
AnswerLet r be the common ratio of the given G.P. $\therefore\text{b}=\text{ar} \text{ and } \text{c}=\text{ar}^2$ Now, $\text{a} + \text{b} + \text{c} = \text{bx}$ $\Rightarrow\text{a}+\text{ar}+\text{ar}^2=\text{arx}$ $\Rightarrow\text{r}^2+(1 - \text{x})\text{r}+1=0$ r is always a real number. $\therefore\text{D}\geq0$ $\Rightarrow(1 -\text{x})^2-4\ge0$ $\Rightarrow\text{x}^2-2\text{x}-3\ge0$ $\Rightarrow(\text{x}-3)(\text{x}+1)\ge0$ $\Rightarrow\text{x}\ge3\text{ or }\text{x}<-1\text{ and }\text{x}\neq3\text{ or }-1$ $[\therefore$ a, b and c are distinct real numbers$]$
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If a, b, c are in G.P., prove that: $\frac{1}{\text{a}^2-\text{b}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{b}^2-\text{c}^2}$
Answer$a, b, c$ are in G.P.a, $b=a r, c=a r^2$
$\text{L.H.S}=\frac{1}{\text{a}^2-\text{b}^2}+\frac{1}{\text{b}^2}$
$=\frac{1}{\text{a}^2-\text{a}^2\text{r}^2}+\frac{1}{\text{a}^2\text{r}^2}$
$=\frac{1}{\text{a}^2}\Big[\frac{1}{1-\text{r}^2}+\frac{1}{\text{r}^2}\Big]$
$=\frac{1}{\text{a}^2}\Bigg[\frac{\text{r}^2+1-\text{r}^2}{\big(1-\text{r}^2\big)\text{r}^2}\Bigg]$
$=\frac{1}{\text{a}^2}\Big[\frac{1}{\text{r}^2-\text{r}^4}\Big]$
$=\frac{1}{(\text{ar})^2-(\text{ar}^2)^2}$
$=\frac{1}{\text{b}^2-\text{c}^2}$
$=\text{R.H.S}$
$\therefore\text{R.H.S}=\text{L.H.S}$
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If $5^{\text {th }}, 8^{\text {th }}$ and $11^{\text {th }}$ term of G.P. are $p, q$ and $s$ respectively, prove that $q^2=p s$.
AnswerLet a be the first term and r be the common ratio of the given G.P. $\therefore\text{p} = 5^{\text{th}} \text{ term}$ $\Rightarrow\text{p}=\text{ar}^4\dots(1)$ $\text{q}=8^\text{th}\text{ term}$ $\Rightarrow\text{q}=\text{ar}^7\dots(2)$ $\text{s}=11^\text{th}$ $\Rightarrow\text{s}=\text{ar}^{10}\dots(3)$ Now, $\text{q}^2=(\text{ar}^7)^2=\text{a}^2\text{r}^{14}$ $\Rightarrow(\text{ar}^4)(\text{ar}^{10})=\text{ps}$ [From (1) and (3)] $\therefore\text{q}^2=\text{ps}$
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If a, b, c are in A.P. and a, b, d are in G.P., then prove that a, a - b, d - c are in G.P.
AnswerHere, $a, b, c$ are in A.P. $2 b=a+c \ldots$...(i) And $a, b, d$ are in G.P., so $b^2=a d \ldots$...(ii)
Now, $(a-b)^2=a^2+b^2-2 a b=a 2+a d-a(a+c)$ Using equation (i) and (ii) $=a^2+a d-a^2-a c=a d-a c(a-b)^2=a(d-c) \frac{(a-b)}{a}=\frac{(d-c)}{(a-b)}$
$\Rightarrow \mathrm{a},(\mathrm{a}-\mathrm{b}),(\mathrm{d}-\mathrm{c})$ are in G.P.
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One side of equilateral triangle is 18 cm. The mid-points of its sides are joined to from another triangle whose mind-points, in turn, are joined to from still another triangle. the process is continued indefinitely. Find the sum of the (i) Perimeters of all the triangles. (ii) Areas of all triangles.
Answer
Side of triangle = 18cm.
AD = BD = 9cm.
DE = BD = 9cm.
$\text{GI} = \text{IF} = \frac{9}{2}\text{cm}.$
Sides of the triangles are $18, 9, \frac{9}{2}\dots$
(i) Sum of perimeters of the equilateral triangle $=\Big(54+27+\frac{27}{2}+\ \dots\Big)$
$=\frac{54}{1-\frac{1}{2}}$
$=54\times2$
Perimeter = 108cm.
(ii) Sum of area of equilateral triangle
$=\bigg[\frac{\sqrt{3}}{4}(18)^2+\frac{\sqrt{3}}{4}(9)^2+\frac{\sqrt{3}}{4}\Big(\frac{9}{2}\Big)^2+\dots\bigg]$
$=\frac{\sqrt{3}}{4}\Big[324+81+\frac{81}{4}+\dots\Big]$
$=\frac{\sqrt{3}}{4}\Bigg[\frac{324}{1-\frac14}\Bigg]$
$=\frac{\sqrt{3}}{4}\Big[\frac{324\times4}{3}\Big]$
$=\sqrt{3}(108)$ View full question & answer→Question 334 Marks
If a, b, c are in G.P., prove that: $(\text{a}+2\text{b}+2\text{c})(\text{a}-2\text{b}+2\text{c})=\text{a}^2+4\text{c}^2$
Answera, b, c are in G.P.$a, b = ar, c = ar^2$
$\text{L.H.S}=({\text{a}+2\text{b}+2\text{c})}{(\text{a}-2\text{ar}+2\text{c})}$
$=\big(\text{a}+2\text{ar}+2\text{ar}^2\big)\big(1-2\text{ar}+2\text{ar}^2\big)$
$=\text{a}^2\big(1+2\text{ar}+2\text{ar}^2\big)\big(1-2\text{r}+2\text{r}^2\big)$
$=\text{a}^2\Big[\big(1+2\text{r}^2\big)^2-\big(2\text{r}\big)^2\Big]$
$=\text{a}^2\big[1+4\text{r}^4+4\text{r}^2-4\text{r}^2\big]$
$=\text{a}^2\big[1+4\text{r}^4\big]$
$=\text{a}^2+4\big(\text{ar}^2\big)^2$
$=\text{a}^2+4\text{c}^2$
$=\text{R.H.S}$
$\therefore\text{R.H.S}=\text{L.H.S}$
View full question & answer→Question 344 Marks
If a, b, c are in A.P. b, c, d are in G.P. and $\frac{1}{\text{c}},\frac{1}{\text{d}},\frac{1}{\text{e}}$ are in A.P., prove that a, c, e are in G.P.
Answera, b, c are in A.P. $\Rightarrow2\text{b}=\text{a}+\text{c }\cdots(\text{i})$ b, c, d are in G.P. $\Rightarrow\text{c}^2-\text{bd }\cdots{\text{ii}}$ $\frac{1}{\text{c}},\frac{1}{\text{d}},\frac{1}{\text{e}}\text{ are in A.P.}$ $\Rightarrow\frac{2}{\text{d}}=\frac{1}{\text{c}}+\frac{1}{\text{e}}\cdots(\text{iii})$ We need to prove that a, b, c are in G.P. $\Rightarrow\text{c}^2=\text{ae}$ Now, $\text{c}^2=\text{bd}=2\text{b}\times\frac{\text{d}}{2}$ $\Rightarrow\text{c}^2=(\text{a}+\text{c})\times\frac{\text{ce}}{\text{c}+\text{e}}$ $\Big[\because\frac{2}{\text{d}}=\frac{\text{e}+\text{c}}{\text{ce}}\Big]$ $\Rightarrow\text{c}^2=\frac{(\text{a}+\text{c})\text{ce}}{\text{c}+\text{e}}$ $\Rightarrow\text{c}^2(\text{c}+\text{e})=\text{ace}+\text{c}^2\text{e}$ $\Rightarrow\text{c}^3+\text{c}^2\text{e}=\text{ace}+\text{c}^2\text{e}$ $\Rightarrow\text{c}^3=\text{ace}$ $\Rightarrow\text{c}^2=\text{ae}$ Hence proved.
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Find the sum of the following series: 9 + 99 + 999 + ... to n terms.
Answer$9 + 99 + 999 + ...\text{ to n terms}$ This can be written as $\big(10-1\big)+\big(100-1\big)+\big(1000-1\big)+\dots\text{n terms}$ $\big(10+10^2+10^3+\dots\text{n terms}\big)-\text{n}$ $\Rightarrow\text{S}_\text{n}=\frac{\text{a}({\text{r}^\text{n}-1})}{\text{10}-1},\text{a}=10,\text{r}=10,\text{n}=\text{n}$ $=\frac{10(10^\text{n}-1)}{10-1}-\text{n}$ $=\frac{10}{9}(10^\text{n}-1)-\text{n}$ $=\frac19\big[10^{\text{n}+1}-10-9\text{n}\big]$ $=\frac{1}{9}\big[10^{\text{n}+1}-9\text{n}-10\big]$
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If a, b, c, are in G.P., prove that the following are also in G.P. $\text{a}^2,\text{b}^2,\text{c}^2$
Answera, b, c are in G.P.$\Rightarrow\text{b}^2=\text{ac }\cdots{(1)}$
$\big(\text{b}^2\big)=\big(\text{ac}\big)^2$
$\big(\text{b}^2\big)^2=\text{a}^2\text{c}^2$
$\Rightarrow\text{a}^2,\text{b}^2,\text{c}^2\text{ are in G.P.}$
View full question & answer→Question 374 Marks
The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers A.P. Find the numbers.
AnswerLet the number are: $\frac{\text{a}}{\text{r}},\text{a}\text{ and }\text{ar}.$ Then, $\frac{\text{a}}{\text{r}}+\text{a}+\text{ar}=14$ Again the numbers a + 1, ar + 1 and ar2 - 1 are in A.P. $\therefore 2(\text{a}+1)=(\text{ar}-1)+\Big(\frac{\text{a}}{\text{r}}+1\Big)$ $ 2(\text{a}+1)=\text{ar}+\frac{\text{a}}{\text{r}}$ $2(\text{a}+1)=14-\text{a}$ $3\text{a}=12$ $\text{a}=4$ Now we have $\frac{4}{\text{r}}+4+4\text{r}=14$ $2-5\text{r}+2\text{r}^2=0$ $2\text{r}^2-4\text{r}-\text{r}+2=0$ $2\text{r}(\text{r}-2)-1(\text{r}-2)=0$ $(\text{r}-2)(2\text{r}-1)=0$ $\text{r}=2,\frac12$ Thus the numbers are: 2, 4, 8 or 8, 4, 2.
View full question & answer→Question 384 Marks
Evaluate the following:$\sum_\limits{\text{k}=1}^{\text{n}}(2^\text{k}+3^{\text{k}-1})$
Answer$\sum_\limits{\text{k}=1}^{\text{n}}(2^\text{k}+3^{\text{k}-1})$ $=(2+3^0)+(2^2+3)+(2^3+3^2)+\ \dots\ +(2^\text{n}+3^{{\text{n}-}1})$ $(2+2^2+2^3+\ \dots\ +2^{\text{n}})+(3^0+3^1+3^2+\ \dots\ +3^{\text{n}-1})$ $=\text{S}_\text{n}+\text{S}_\text{m}$ $\text{S}_\text{n}\Rightarrow\text{a}=2,\text{n}=\text{n},\text{r}=\frac{2^2}{2}=2$ $\text{S}_\text{n}=\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}=\frac{2(2^\text{n}-1)}{2-1}=2(2^\text{n}-1)$ Also, $\text{S}_\text{m}=\text{S}_{\text{n}-1}$ $\text{a}=1,\text{r}=3,\text{n}=\text{n}-1$ $\text{S}_{\text{n}-1}=\frac{1(3^{\text{n}-1}-1)}{3-1}=\frac12(3^{\text{n}}-1)$ $\therefore\sum_\limits{\text{k}-1}^\text{n}(2^{\text{k}}+3^{\text{k}-1})=2(2^\text{n}-1)+\frac12(3^\text{n}-1)$ $=\frac12\big[2^{\text{n}+2}+3^\text{n}-4-1\big]$ $=\frac12\big[2^{\text{n}+2}+3^\text{n}-5\big]$
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The sum of first three term of a G.P. is $\frac{13}{12}$ and their product is $-1.$ Find the G.P.
AnswerLet the first three terms of G.P. are $\frac{\text{a}}{\text{r}},\text{a},\text{ar}$ Here, $\frac{\text{a}}{\text{r}}+\text{a}+\text{ar}=\frac{13}{12}\cdots(\text{i})$ and $\frac{\text{a}}{\text{r}}\times\text{a}\times\text{ar}=-1$ $\Rightarrow\text{a}^3=-1$ $\Rightarrow\text{a}=-1$ Put a = -1 in equition (i), $\frac{-1}{\text{r}}+(-1)-\text{r}=\frac{13}{12}$. $\Rightarrow-1-\text{r}-\text{r}^2=\frac{13}{12}\text{r}$ $\Rightarrow-12-12\text{r}=12\text{r}^2=13\text{r}$ $\Rightarrow12\text{r}^2+12\text{r}+13\text{r}+12=0$ $\Rightarrow12\text{r}^2+25\text{r}+12=0$ $\Rightarrow12\text{r}^2+16\text{r}+9\text{r}+12=0$ $\Rightarrow4\text{r}(3\text{r}+4)+3(3\text{r}+4)=0$ $\Rightarrow(4\text{r}+3)(3\text{r}+4)=0$ $\text{r}=\frac{-3}{4},\frac{-4}{3}$ So, Required G.P. is, $\frac{4}{3},-1,\frac{3}{4},\ ...$ or $\frac{3}{4},-1,\frac{4}{3},\ ...$
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If $S_1, S_2, \ldots, S_n$ are the sums of $n$ terms of $n$ G.P.'s whose first terms is 1 in each and common ratio are $1,2,3, \ldots, n$ respectively, then prove that $S_1,+S_2+2 S_3+3 S_4+\ldots(n-1) S_n=1^n+2^n+3^n+\ldots+n^n$.
Answer$S_1, S_2, \ldots, S_n$ are the sum of $n$ terms of G.P. $a=1, r=1,2,3, \ldots, n$ Then, $S_1+S_2+2 S_3+3 S_4+\ldots+(n-1) S_n$
$\frac{1(1^\text{n}-1)}{1-1}+\frac{1(2^\text{n}-1)}{2-1}+\frac{2(3^\text{n}-1)}{3-1}+\ ... \ + (\text{n}-1)1\Big(\frac{1^\text{n}-1}{1-1}\Big)$
$=2^\text{n}-1+23^\text{n}-1+3.4^\text{n}-1+\ ...$
$=2^\text{n}+3^\text{n}+4^\text{n}+\ ...\ +\text{n}^\text{n}$
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If a, b, c, d are in G.P., prove that: $(\text{a}+\text{b}+\text{c}+\text{d})^2=(\text{a}+\text{b}^2)+2(\text{b}+\text{c})^2+(\text{c}+\text{d})^2$
Answera, b and c are in G.P.$\therefore\text{b}^2=\text{ac }\cdots(1)$
$\text{L.H.S}=({\text{a}+\text{b}+\text{c}+\text{d})^2}$
$=(\text{a}+\text{b})^2+2(\text{a}+\text{b})(\text{c}+\text{d})+(\text{c}+\text{d})^2$
$=(\text{a}+\text{b})^2+2(\text{ac}+\text{ad}+\text{bc}+\text{bd})+(\text{c}+\text{d})^2$
$=(\text{a}+\text{b})^2+2(\text{b}^2+\text{bc}+\text{bc}+\text{c}^2)+(\text{c}+\text{d})^2$ $[\text{Using (1)}]$
$=(\text{a}+\text{b})^2+2(\text{b}+\text{c})^2+(\text{c}+\text{d})^2$
$=\text{R.H.S}$
$\therefore\text{R.H.S}=\text{L.H.S}$
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In a G.P. the $3^{rd}$ term is 24 and the 6th term is 192. Find the 10th term.
AnswerLet the first term is a and the common ratio of is r. Then, $\text{ar}^2=24\dots(1)$ and $\text{ar}^5=192\dots(2)$ (2) ÷ (1), we get $\frac{\text{ar}^5}{\text{ar}^2}=\frac{192}{24}$ $\text{r}^3=8$ $\text{r}=2$ Now, $\text{ar}^2=24$ $\text{a}\cdot2^2=24$ $\text{a}=6$ Thus the 10th term will be: $\text{ar}^9=6\cdot2^9=3072$
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The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them, the results are in A.P. Find the numbers.
AnswerLet the number in G.P. are $\frac{\text{a}}{\text{r}},\text{a}\text{ and }\text{ar}.$ So, $\frac{\text{a}}{\text{r}}\times\text{a}\times\text{ar}=216$ $\Rightarrow\text{a}^3=216$ $\Rightarrow\text{a}=6$ Again also given, $\frac{\text{a}}{\text{r}}+2,\text{a}+8,\text{ar}+6$ are in A.P. $ 2(\text{a}+8)=\Big(\frac{\text{a}}{\text{r}}+2\Big)+(\text{ar}+6)$ $\Rightarrow2(6+8)=\Big(\frac{6+2\text{r}}{\text{r}}\Big)+6\text{r}+6$ $\Rightarrow28\text{r}=6+2\text{r}+6\text{r}^2+6\text{r}$ $\Rightarrow6\text{r}^2-20\text{r}+6=0$ $\Rightarrow6\text{r}^2-18\text{r}-2\text{r}+6=0$ $\Rightarrow6\text{r}(\text{r}-3)-2(\text{r}-3)=0$ $\Rightarrow(\text{r}-3)(6\text{r}-2)=0$ $\text{r}=3,\text{r}=\frac{1}{3}$ So, Required G.P. is 18, 6, 2, ... Or, 2, 6, 18 ...
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If $\text{x}^{\text{a}}=\text{x}^{\frac{\text{b}}{2}}\text{z}^{\frac{\text{b}}{2}}=\text{z}^{\text{c}},$ then prove that $\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}$ are in A.P.
Answer$\text{x}^{\text{a}}=\text{x}^{\frac{\text{b}}{2}}\text{z}^{\frac{\text{b}}{2}}=\text{z}^{\text{c}}=\lambda(\text{ say})$ $\text{x}=\lambda^{\frac{1}{\text{a}}},\text{z}=\lambda^{\frac{1}{\text{c}}}$ $\text{x}^{\frac{\text{b}}{2}}\times\text{z}^{\frac{\text{b}}{2}}=\lambda$ $\lambda^{\frac{1}{\text{a}}\big(\frac{\text{b}}{2}\big)}\times\lambda^{\frac{\text{b}}{2}\times\frac{1}{\text{c}}}=\lambda$ $\lambda^{\frac{\text{b}}{2\text{a}}+\frac{\text{b}}{2\text{c}}}=\lambda^1$ $\frac{\text{b}}{2\text{a}}+\frac{\text{b}}{2\text{c}}=1$ $\frac{1}{\text{a}}+\frac{1}{\text{c}}=\frac{2}{\text{b}}$ $\Rightarrow\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}\text{ are in A.P.}$
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Prove that the product of n geometric means between two quantities is equal to the nth power of a geometric mean of those two quantities.
AnswerLet $G_1, G_2, G_3, G_4, ..., G_n$ be n G.M. is between a and b. Then, a, $G_1, G_2, G_3, G_4, ..., G_n$, b is a G.P. Let r be the common ratio. $\because\text{b}=\text{a}_{\text{n}+2}=\text{ar}^{(\text{n}+1)}$
$\Rightarrow\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{(\text{n}+1)}}$
$\therefore\text{G}_1=\text{a}_2=\text{ar}$
$\text{G}_2=\text{a}_3=\text{ar}^2$
$\text{G}_3=\text{a}_4=\text{ar}^3$ . . . $\text{G}_\text{n}=\text{a}_{(\text{n}+1)}=\text{ar}^{\text{n}}$ Also, let G be the G.M. between a and b. $\therefore\text{G}^2=\text{ab}$ Now, $G_1 \times G_2 \times G_3 \times G_4 \times G_5 \times ... \times G_n = ar \times ar^2 \times ar^3 \times ar^4 \times ... \times ar^n =\text{a}^{\text{n}}\times\text{r}^{(1+2+3+4+ ... +\text{n})}$
$=\text{a}^{\text{n}}\times\text{r}^{\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)}$
$=\text{a}^{\text{n}}\times\Bigg[\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{(\text{n}+1)}}\Bigg]^{\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)}$
$=\text{a}^{\text{n}}\times\Big(\frac{\text{b}}{\text{a}}\Big)^\frac{\text{n}}{2}$
$=\text{a}^{\frac{\text{n}}{2}}\times\text{b}^{\frac{\text{n}}{2}}$
$=(\text{ab})^{\frac{\text{n}}{2}}$
$=\Big(\sqrt{\text{ab}}\Big)^{\text{n}}$
$=\text{G}_\text{n}$
$\therefore\text{G}_1\times\text{G}_2\times\text{G}_3\times\text{G}_4\times\ ...\times\text{G}_\text{n}=\text{G}^{\text{n}}$
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Find three numbers in G.P. whose Product is 729 and the sum of their products in pairs is 819.
AnswerLet the number in G.P. are $\frac{\text{a}}{\text{r}},\text{a}\text{ and }\text{ar}.$ Here, $\frac{\text{a}}{\text{r}}\times\text{a}\times\text{ar}=729$ $\Rightarrow\text{a}^3=729$ $\Rightarrow\text{a}=9$ And, $\Big(\frac{\text{a}}{\text{r}}\times\text{a}\Big)+(\text{a}\times\text{ar})+\Big(\frac{\text{a}}{\text{r}}\times\text{ar}\Big)=819$ $\Rightarrow\frac{81}{\text{r}}+81\text{r}+81=819$ $\Rightarrow\frac{9}{\text{r}}+9\text{r}+9=91$ $\Rightarrow9+9\text{r}^2+9\text{r}=91\text{r}$ $\Rightarrow9\text{r}^2-81\text{r}-\text{r}+9=0$ $\Rightarrow9\text{r}(\text{r}-9)-1(\text{r}-9)=0$ $\text{r}=9,\frac{1}{9}$ So, required G.P. are 81, 9, 1, ... Or, 1, 9, 81, ...
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Find the sum of 2n terms of the series whose every even term is 'a' times the term before it and every odd term is 'c' times the term before it, the first term being unity.
Answer$\text { Let the series be } a_1+a_2+a_3+\ldots+a_{2 n} \text { It is given that } a_1=1, a_2=a_3=a c, a_4=a^2 c, a_5=a^2 c^2, \ldots . \therefore \text { Sum of } 2 n \text { term } a_1$
$+a_2+a_3+\ldots+a_{2 n}=1+a+a c+a^2 c+a^2 c^2+\ldots+2 n \text { term }=(1+a)+a c(1+a)+a^2 c^2(1+a)+\ldots+n \text { term }$
$=(1 +\text{a})\frac{\big(1-(\text{ac})^{\text{n}}\big)}{1-\text{ac}}$ $=(\text{a}+1)\frac{\big((\text{ac})^{\text{n}}-1\big)}{\text{ac}-1}.$
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Find the rational number having the following decimal expansions:$3.\overline{52}$
Answer$3.\overline{52}=3+0.52222\ \dots$$=3+0.5+0.02+0.002+0.0002+\ \dots$
$=3.5+\frac{2}{10^2}+\frac{2}{10^3}+\frac{2}{10^4}+\ \dots$
$=3.5+\frac{2}{10^2}\Big(1+\frac{1}{10}+\frac{1}{10^2}+\ \dots\Big)$
$=\frac{35}{10}+\frac{2}{100}\Bigg(\frac{1}{1-\frac{1}{10}}\Bigg)$
$=\frac{35}{10}+\frac{2}{100}\times\Big(\frac{10}{9}\Big)$
$=\frac{35}{10}+\frac{2}{90}$
$=\frac{315+2}{90}$
$3.\overline{52}=\frac{317}{90}$
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Find the rational number having the following decimal expansions:$0.\overline{68}$
AnswerThe rational number can be written as: $0.\overline{68}=0.6+0.08+0.008+0.0008+\ \dots\infty$ $=\frac35+8[0.01+0.001+0.0001+0.00001+\ \dots\infty]$ $=\frac{3}{5}+8\Big[\frac{1}{100}+\frac{1}{100}+\dots\infty\Big]$ This is an infinite G.P. with first term $\frac{1}{100}$ and common ratio $\frac{1}{10}$ $=\frac{3}{5}+8\times\frac{1}{100}\times\frac{1}{1-\frac{1}{10}}$ $=\frac{3}{5}+\frac{4}{45}$ $=\frac{31}{45} $
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If (a - b), (b - c), (c - a) are in G.P., then prove that: $\big(\text{a}+\text{b}+\text{c}\big)^2=3\big(\text{ab}+\text{bc}+\text{ca}\big)$
Answer$(\text{a} - \text{b}), (\text{b} - \text{c}), (\text{c} - \text{a}) \text{ are in G.P.}$ $(\text{b} - \text{c})^2 = (\text{a} - \text{b}) (\text{c} - \text{a})$ $\text{b}^2 + \text{c}^2 - 2\text{bc} = \text{ac} - \text{a}^2 - \text{bc} + \text{ab}$ $\text{b}^2 + \text{c}^2 + \text{a}^2 = \text{ac} + \text{bc} + \text{ab}\cdots(\text{i})$ Now, $(\text{a}+\text{b}+\text{c})^2=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}$ $=\text{ac}+\text{bc}+\text{ab}+2\text{ab}+2\text{bc}+2\text{bc}+2\text{ca}$ [Using equation (1)] $=3\text{ab}+3\text{bc}+3\text{ca}$ $(\text{a}+\text{b}+\text{c})^2=3(\text{ab}+\text{bc}+\text{ca})$
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If $S_1, S_2, S_3$, be respectively the sums of n, 2n, 3n terms of a G.P., then prove that $\text{S}^2_1+\text{S}^2_2=\text{S}_1(\text{S}_2+\text{S}_3).$
Answer$S_1$ = sum of n terms, S1 = sum of 2n terms, S1 = sum of 3n terms. Then, $\text{S}_1^2+\text{S}_2^2$
$=(\text{S}_\text{n})^2+(\text{S}_\text{2n})^2$
$=\Big(\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}\Big)^2+=\Big(\frac{\text{a}(1-\text{r}^\text{2n})}{1-\text{r}}\Big)^2$
$=\frac{\text{a}^2}{(1-\text{r})^2}\Big[\big(1-(\text{r})^\text{n})^2+(1-\text{r}^{2\text{n}})\Big]$
$=\frac{\text{a}^2}{(1-\text{r})^2}\Big[1+\text{r}^{2\text{n}}-2\text{r}^\text{n}+1+\text{r}^{4\text{n}}-2\text{r}^{2\text{n}}\Big]$
$=\frac{\text{a}^2}{(1-\text{r})^2}\Big[2-\text{r}^{2\text{n}}-2\text{r}^\text{n}+\text{r}^{4\text{n}}\Big]\cdots(\text{i})$ Also, $\text{S}_1(\text{S}_2+\text{S}_3)$
$=\frac{\text{a}^2}{(1-\text{r})^2}\Big(\frac{\text{a}(1-\text{r}^{2\text{n}})}{1-\text{r}}+\frac{\text{a}(1-\text{r}^{3\text{n}})}{1-\text{r}}\Big)$
$=\frac{\text{a}^2}{(1-\text{r})^2}\big[(1-\text{r})^\text{n}(1-\text{r}^{2\text{n}})+(1-\text{r}^{\text{n}})(1-\text{r}^{3\text{n}})\big]$
$=\frac{\text{a}^2}{(1-\text{r})^2}\big[1-\text{r}^{2\text{n}}-\text{r}^\text{n}+\text{r}^{3\text{n}}-\text{r}^{3\text{n}}-\text{r}^{\text{n}}+1+\text{r}^{4\text{n}}\big]$
$=\frac{\text{a}^2}{(1-\text{r})^2}\big[2-\text{r}^{2\text{n}}-2\text{r}^\text{n}+\text{r}^{4\text{n}}\big]\cdots(\text{ii})$
$(\text{i})=(\text{ii})\text{ Hence, }\text{S}^2_1+\text{S}^2_2=\text{S}_1(\text{S}_2+\text{S}_3)$
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If a, b, c, are in G.P., prove that: $\big(\text{a}^2-\text{b}^2\big),\big(\text{b}^2-\text{c}^2\big),\big(\text{c}^2-\text{d}^2\big)\text{ are in G.P.}$
Answera, b, c, d are in G.P. $\therefore\text{b}^2=\text{ac}$ $\text{ad}=\text{bc}$ $\text{c}^2=\text{bd}\cdots(1)$ Now, $\big(\text{b}^2-\text{c}^2\big)^2=\big(\text{b}^2\big)-2\text{b}^2\text{c}^2+\big(\text{c}^2\big)^2$ $\Rightarrow\big(\text{b}^2-\text{c}^2\big)^2=\big(\text{ac}\big)^2-\text{b}^2\text{c}^2-\text{b}^2\text{c}^2+\big(\text{bd}\big)^2$ [Using (1)] $\Rightarrow\big(\text{b}^2-\text{c}^2\big)^2=\text{a}^2\text{c}^2-\text{b}^2\text{c}^2-\text{a}^2\text{d}^2+\text{b}^2\text{d}^2$ [Using (1)] $\Rightarrow\big(\text{b}^2-\text{c}^2\big)^2=\text{c}^2(\text{a}^2-\text{b}^2\big)-\text{d}^2\big(\text{a}^2-\text{b}^2\big)$ $\Rightarrow\big(\text{b}^2-\text{c}^2\big)^2=\big(\text{a}^2-\text{b}^2\big)\big(\text{c}^2-\text{d}^2\big)$ $\therefore\big(\text{a}^2-\text{b}^2\big),\big(\text{b}^2-\text{c}^2\big)\text{ and }\big(\text{c}^2-\text{d}^2\big)\text{ are also in G.P.}$
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A Person has 2 parents, 4 grandparents, 8 great parents, and so on. Find the number of his ancestors during the generation preceding his own.
AnswerTo find number of ancestors, we will find the sum of $2, 2^2, 2^3$, ... Number of ancestors $=\frac{2(2^{10}-1)}{2-1}$ $=2(1024-1)$ $=2\times1023$ $=2046$
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If the $p^{th}$ and $q^{th}$ terms of a G.P. and q and p respectively, show that $(p + q)^{th}$ term is $\Big(\frac{\text{q}^\text{p}}{\text{p}^\text{q}}\Big)^\frac{1}{\text{p}-\text{q}}.$
Answer$n^{th}$ term of $G.P. = ar^{x-1} p^{th}$ term =$q = a.r^{p-1} q^{th}$ term = $p = a.r^{q-1}$^ $\frac{\text{q}}{\text{p}}=\text{r}^{\text{r}-4}$
$\text{r}=\Big(\frac{\text{q}}{\text{p}}\Big)^\frac{1}{\text{r}-\text{q}}$
$\text{a}=\text{p}\Big(\frac{\text{p}}{\text{q}}\Big)^\frac{1-\text{q}}{\text{p}-\text{q}}$
$\text{p}+\text{q}^\text{th}\text{term}=\text{p}\Big(\frac{\text{q}}{\text{p}}\Big)^{\frac{1-\text{q}}{\text{p}-\text{q}}}\Big(\frac{\text{q}}{\text{p}}\Big)^{\frac{\text{p}+\text{q}-1}{\text{p}-\text{q}}}$
$=\text{p}\Big(\frac{\text{q}}{\text{p}}\Big)^{\frac{1-\text{q}+\text{p}+\text{q}-1}{\text{p}-\text{q}}}$
$=\text{p}\Big(\frac{\text{q}}{\text{p}}\Big)^{\frac{\text{p}}{\text{p}-\text{q}}}$
$=\frac{\text{q}^\frac{\text{p}}{\text{p}-\text{q}}}{\text{p}^{\frac{\text{q}}{\text{p}-\text{q}}-1}}$
$=\frac{\text{q}^\frac{\text{p}}{\text{p}-\text{q}}}{\text{p}^{\frac{\text{q}}{\text{p}-\text{q}}}}$
$=\Big(\frac{\text{q}^\text{p}}{\text{p}^\text{q}}\Big)^\frac{1}{\text{p}-\text{q}}$
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If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively, then obtain the quadratic equation.
AnswerLet the roots of the quadratic equation be a and b. AM = 8 $\therefore\ \frac{\text{a}+\text{b}}{2}=8$ $\Rightarrow\text{a}+\text{b}=16\ \cdots(\text{i})$ Also, G = 5 $\Rightarrow\sqrt{\text{ab}}=5$ $\Rightarrow\text{ab}=5^2$ $\Rightarrow\text{ab}=25\ \cdots(\text{ii})$ Now, the quadratic equition is given by $\text{x}^2-(\text{a}+\text{b})\text{x}+\text{ab}=0$ $\Rightarrow\text{x}^2-16\text{x}+25=0$
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Show that in an infinite G.P. with common ratio $\text{r}\big(|\text{r}|<1\big),$ each terms bears a constant ratio to the sum of all terms that follow it.
AnswerLet a be first term and r be common ratio of G.P. Here, $\frac{\text{a}_\text{n}}{\big(\text{a}_{\text{n}+1}+\text{a}_{\text{n}+2}+\ \dots\infty\big)}=\frac{\text{ar}^{\text{n}-1}}{\text{ar}^{\text{n}}+\text{ar}^{\text{n}+1}+\ \dots}$ $=\frac{\text{ar}^{\text{n}-1}}{\text{ar}^{\text{n}}\big(1+\text{r}+\text{r}^2+\ \dots\infty\big)}$ $=\frac{\text{ar}^{\text{n}-1}}{\text{ar}^{\text{n}}\Big(\frac{1}{1-\text{r}}\Big)}$ $=\Big(\frac{1-\text{r}}{\text{r}}\Big)$ Since r is a constant, so $\Big(\frac{\text{a}_\text{n}}{\text{a}_{\text{n}+1}+\text{a}_{\text{n}+2}+\ \dots\infty}\Big)=\text{k}\ (\text{constant})$ Such that $\text{k}=\Big(\frac{1-\text{r}}{\text{r}}\Big)$
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