3 Marks Question

Question 13 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.

View full question & answer→

Question 23 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, t₁₆ = 275, n = 16 and S₁₆ = 8000
$\therefore$ t₁₆ = a + (16 - 1)(-d)
⇒ 275 = a - 15d ....(1)
Also, $\text{S}_{16}=\frac{16}{2}[2\text{a}+(\text{n}-1)(-\text{d})]$
⇒ 8000 = 8[2a + (16 - 1)(-d)]
⇒ 1000 = 2a - 15d ....(2)
Solving Eqs. (1) and (2), we get a = 725
Hence, first place team recieves Rs. 725.

View full question & answer→

Question 33 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
$\text{a}_6=\text{ar}^{6-1}=60\times\Big(\frac{1}{2}\Big)^5=\frac{60}{32}=\frac{15}{8}\text{cm}$

View full question & answer→

Question 43 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 = a₁₀

= a + (n - 1)d

= 5200 + (10 - 1) × 320

= 5200 + 9 × 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 = S₁₂

$=\frac{12}{2}[2\times5200+(12-1)3202]$

= 6[10400 + 11 × 320]

= 6[10400 + 3520]

= 6 × 13920 = Rs. 83520

View full question & answer→

Question 53 Marks

If a₁, a₂, a₃, ..., 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., $\forall\ \text{a}_\text{i}>0$
$\therefore$ a₁- a₂ = a₂ - a₃ = .... = 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 = a₁ + (n - 1)d)
$=\frac{\text{n}-1}{\sqrt{\text{a}_1}+\sqrt{\text{a}_\text{n}}}$

View full question & answer→

Question 63 Marks

If A is the arithmetic mean and G₁ , G₂ 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, G₁ and G₂ are geometric means between a and b, the a, G₁, G₂, 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 $\text{G}_2=\text{ar}^2=\text{a}\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{2}{3}}=\text{a}^{\frac{1}{3}}\text{b}^{\frac{2}{3}}$
$\therefore\ \frac{\text{G}^2_1}{\text{G}_2}+\frac{\text{G}^2_2}{\text{G}_1}=\frac{\text{G}^3_1+\text{G}^3_2}{\text{G}_1\text {G}_2}=\frac{\text{a}^2\text{b}+\text{ab}^2}{\text{ab}}=\text{a}+\text{b}=2\text{A}$

View full question & answer→

Question 73 Marks

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

	Column I		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

	Column I		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.

Solution:

$4,1,\frac{1}{4},\frac{1}{16}$

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.

View full question & answer→

Question 83 Marks

Find the r^th term of an A.P. sum of whose first n terms is 2n + 3n² .
[Hint: an = S_n– S_n– 1]

Answer

Given that S_n = 2n + 3n²
⇒ S₁ = 2n + 3n²
⇒ S₂ = 2 × 1 + 3(1)² = 5
⇒ S₃ = 2 × 3 × 9 = 33
$\therefore$ S₁ = a₁ = 5
S₂ - S₁ = a₂ = 16 - 5 = 11
$\therefore$ d = a₂ - a₁ = 11 - 5
Now T_r = a₁ + (r - 1)d
= 5 + (r - 1)6 = 5 + 6r - 6 = 6r - 1
Hence, the required rth term is 6r - 1.

View full question & answer→

MATHS 3 Marks Question

Take a timed test