3 Marks Question

Question 13 Marks

Match the questions given under Column I with their appropriate answers given under the Column $II.$

	$\text{Column I} $		$\text{Column II} $
$(a)$	$4,1,\frac{1}{4},\frac{1}{16}$	$(i)$	$A.P.$
$(b)$	$2, 3, 5, 7$	$(ii)$	$Squence$
$(c)$	$13, 8, 3, -2, -7$	$(iii)$	$G.P.$

Answer

	$\text{Column I}$		$\text{Column II}$
$(a)$	$4,1,\frac{1}{4},\frac{1}{16}$	$(i)$	$G.P.$
$(b)$	$2, 3, 5, 7$	$(ii)$	$Squence$
$(c)$	$13, 8, 3, -2, -7$	$(iii)$	$A.P.$

$4,1,\frac{1}{4},\frac{1}{16}$
Here, $\frac{\text{a}_2}{\text{a}_1}=\frac{1}{4},\frac{\text{a}_3}{\text{a}_2}=\frac{\frac{1}{4}}{1}=\frac{1}{4}$ and $\frac{\text{a}_4}{\text{a}_3}=\frac{\text{a}_4}{\text{a}_3}=\frac{\frac{1}{16}}{\frac{1}{4}}=\frac{1}{4}$
Hence, it is $G.P.$
$2, 3, 5, 7$
Here$, a2 - a1 = 3 - 2 = 1 a3 - a2 = 5 - 3 = 2 \therefore\ \text{a}_2-\text{a}_1\neq\text{a}_3-\text{a}_2$ Hence$,$ it is not $A.P. \frac{\text{a}_2}{\text{a}_1}=\frac{3}{2},\frac{\text{a}_3}{\text{a}_2}==\frac{5}{3}$
So $,\frac{3}{2}\neq\frac{5}{3}$
So$,$ it is not $G.P.$ Hence, it is squence.
$13, 8, 3, -2, -7$
Here$, a3 - a1 = 8 - 13 = -5 a3 - a2 = 3 - 8 = -5 So, a2 - a1 = a3 - a2 = -5$
So $,$ it is an $A.P.$

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Question 23 Marks

A man accepts a position with an initial salary of $Rs. 5200$ per month. It is understood that he will receive an automatic increase of $Rs. 320$ in the very next month and each month thereafter.

Find his salary for the tenth month.
What is his total earnings during the first year?

Answer

The man gets a fixed increment of $Rs. 320$ each month. Therefore$,$ this forms an $A.P$. whose$,$
First term$, a = 5200$ and Common difference$, d = 320$
Salary for $10^{th}$ month will be given by an$,$ where $n = 10.$
$\therefore$ Total earning $= a_{10}$
$= a + (n - 1)d$
$= 5200 + (10 - 1) \times 320$
$= 5200 + 9 \times 320$
$= 5200 + 2880$
$= Rs. 8080$
Total earnings during the first year is equal to the sum of $12$ terms of the $A.P.$
$\therefore$ Total earnings $= S_{12}$
$=\frac{12}{2}[2\times5200+(12-1)3202]$
$= 6[10400 + 11 \times 320]$
$= 6[10400 + 3520]$
$= 6 \times 13920$
$= Rs. 83520$

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Question 33 Marks

A man saved Rs. 66000 in 20 years. In each succeeding year after the first year he saved Rs. 200 more than what he saved in the previous year. How much did he save in the first year?

Answer

Let Rs. x be saved in first year.
Annual increment = Rs. 200
Which forms an A.P.
First term = a and common difference d =200
n = 20 years
$\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\text{S}_{20}=\frac{20}{2}[2\text{a}+(20-1)200]$
⇒ 66000 = 10[2a + 3800]
⇒ 6600 = 2a + 3800
⇒ 2a = 6600 - 3800
⇒ 2a = 2800
⇒ a = 1400
Hence, the man saved Rs. 1400 in the first year.

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Question 43 Marks

Find the $r^{th}$ term of an $A.P.$ sum of whose first n terms is $2n + 3n^2 $.
$[$Hint: $an = S_n– S_n– 1]$

Answer

Given that $S_n = 2n + 3n^2$
$\Rightarrow S_1 = 2n + 3n^2$
$\Rightarrow S_2 = 2 \times 1 + 3(1)^2 = 5$
$\Rightarrow S_3 = 2 \times 3 \times 9 = 33$
$\therefore S_1 = a_1 = 5$
$S_2 - S_1 = a_2 = 16 - 5 = 11$
$\therefore d = a_2 - a_1 = 11 - 5$
Now $T_r = a_1 + (r - 1)d$
$= 5 + (r - 1)6 = 5 + 6r - 6 = 6r - 1$
Hence$,$ the required rth term is $6r - 1.$

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Question 53 Marks

If $a_1, a_2, a_3, ..., an$ are in$ A.P.,$ where $a_i > 0$ for all $i,$ show that:
$\frac{1}{\sqrt{\text{a}_1}+\sqrt{\text{a}_2}}+\frac{1}{\sqrt{\text{a}_2}+\sqrt{\text{a}_3}}+....+\frac{1}{\sqrt{\text{a}_{\text{n}-1}}+\sqrt{\text{a}_\text{n}}}=\frac{\text{n}-1}{\sqrt{\text{a}_1}+\sqrt{\text{a}_\text{n}}}$

Answer

Given that$, a1 , a2, ....,an$ are in $A.P., \forall\ \text{a}_\text{i}>0$
$\therefore a_1- a_2 = a_2 - a_3 = .... = a_{n- 1} - a_n = -d\ ($constant$)$
Now$,$
$\frac{1}{\sqrt{\text{a}_1}+\sqrt{\text{a}_2}}+\frac{1}{\sqrt{\text{a}_2}+\sqrt{\text{a}_3}}+....+\frac{1}{\sqrt{\text{a}_{\text{n}-1}}+\sqrt{\text{a}_\text{n}}}$
$=\frac{\sqrt{\text{a}_1}-\sqrt{\text{a}_2}}{\text{a}_1-\text{a}_2}+\frac{\sqrt{\text{a}_2}-\sqrt{\text{a}_3}}{\text{a}_2-\text{a}_3}+....+\frac{\sqrt{\text{a}_{\text{n}-1}}-\sqrt{\text{a}_\text{n}}}{\text{a}_{\text{n}-1}-\text{a}_\text{n}} ($rationalizing$)$
$=\frac{\sqrt{\text{a}_1}-\sqrt{\text{a}_2}}{-\text{d}}+\frac{\sqrt{\text{a}_2}-\sqrt{\text{a}_3}}{-\text{d}}+....+\frac{\sqrt{\text{a}_{\text{n}-1}}-\sqrt{\text{a}_\text{n}}}{-\text{d}}$
$=\frac{1}{-\text{d}}\big[\sqrt{\text{a}_1}-\sqrt{\text{a}_\text{n}}\big]$
$=\frac{\text{a}_1-\text{a}_\text{n}}{-\text{d}\big(\sqrt{\text{a}_1}+\sqrt{\text{a}_2}\big)} ($rationalizing$)$
$=\frac{-(\text{n}-1)\text{d}}{-\text{d}\big(\sqrt{\text{a}_1}+\sqrt{\text{a}_\text{n}}\big)}\ (as a_n = a_1 + (n - 1)\ d)$
$=\frac{\text{n}-1}{\sqrt{\text{a}_1}+\sqrt{\text{a}_\text{n}}}$

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Question 63 Marks

In a cricket tournament $16$ school teams participated. $A$ sum of $Rs. 8000$ is to be awarded among themselves as prize money. If the last placed team is awarded $Rs. 275$ in prize money and the award increases by the same amount for successive finishing places$,$ how much amount will the first place team receive?

Answer

Let the first place team get $Rs.$ a as the prize money.
Since award money increases by the same amount for successive finishing places$,$ we get an $A.P.$
Let the constant amount be d.
Here$, t_{16} = 275, n = 16 and S_{16} = 8000$
$\therefore t_{16} = a + (16 - 1)(-d)$
$\Rightarrow 275 = a - 15d ....(1)$
Also, $\text{S}_{16}=\frac{16}{2}[2\text{a}+(\text{n}-1)(-\text{d})]$
$\Rightarrow 8000 = 8[2a + (16 - 1)(-d)]$
$\Rightarrow 1000 = 2a - 15d ....(2)$
Solving Eqs. $(1)$ and $(2),$ we get $a = 725$
Hence$,$ first place team recieves $Rs. 725.$

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Question 73 Marks

If $A$ is the arithmetic mean and $G_1 , G_2$ be two geometric means between any two numbers$,$ then prove that:
$2\text{A}=\frac{\text{G}_1^2}{\text{G}_2}+\frac{\text{G}_2^2}{\text{G}_2}$

Answer

Let the numbers be $a$ and $b.$
Then, $\text{A}=\frac{\text{a}+\text{b}}{2}$ or $\text{2A = a + b .... (1)}$
Also$, G_1$ and $G_2$ are geometric means between $a$ and $b,$ the $a, G_1, G_2, b$ are in $G.P.$
Let $r$ be the common ratio.
Then$, \text{b}=\text{ar}^{4-1}=\text{ar}^3\Rightarrow\frac{\text{b}}{\text{a}}=\text{r}^3\Rightarrow\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big )^{\frac{1}{3}}$
$\therefore\ \text{G}_1=\text{ar}^2=\text{a}\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{3}}=\text{a}^{\frac{2}{3}}\text{b}^{\frac{1}{3}}$
and $\text{G}_2=\text{ar}^2=\text{a}\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{2}{3}}=\text{a}^{\frac{1}{3}}\text{b}^{\frac{2}{3}}$
$\therefore\ \frac{\text{G}^2_1}{\text{G}_2}+\frac{\text{G}^2_2}{\text{G}_1}=\frac{\text{G}^3_1+\text{G}^3_2}{\text{G}_1\text {G}_2}=\frac{\text{a}^2\text{b}+\text{ab}^2}{\text{ab}}=\text{a}+\text{b}=2\text{A}$

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Question 83 Marks

A side of an equilateral triangle is 20cm long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. This process is continued for third, fourth, fifth, triangles. Find the perimeter of the sixth inscribed equilateral triangle.

Answer

Let the given equilateral triangle be $\triangle\text{ABC}$ with each side of 20cm.
By joining the mid-points of this triangle, we get another equilateral triangle of side equal to half of the length of side of $\triangle\text{ABC.}$
Continuing in this way, we get a set of equilateral triangles with side equal to half of the side of the previous triangle.
Now,
Perimeter of first triangle = 20 × 3 = 60cm;
Perimeter of second triangle = 10 × 3 = 30cm;
Perimeter of third triangle = 5 × 3 = 15cm;
Clearly 60, 30, 15 ...., from a G.P. with a = 60 and $\text{r}=\frac{30}{60}=\frac{1}{2}.$
We have, tofind perimeter of sixth incribed triangle i.e., we have to fiind the sixth term of the G.P.
$\therefore$ Perimeter of sixth incribed triangle
$\text{a}_6=\text{ar}^{6-1}=60\times\Big(\frac{1}{2}\Big)^5=\frac{60}{32}=\frac{15}{8}\text{cm}$

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