11 - Arithmetic Mean (Algebra)

In this post, you can check the solutions to the exercises of Arithmetic Mean of Algebra chapter of Readmore Optional Mathematics by DR Simkhada. It consists of complete answers to all odd questions mentioned in the textbook.

Before starting, make sure to check the following note of Determinant.

Things to remember:

  1. The first term of the sequence is denoted by $t_1$ or a.
  2. The common difference of the sequence is denoted by d.
  3. The general formula for AM or m is $m = a + nd$ where n refers to the number of means.
  4. Total number of means in between two numbers is always less than the total number of terms in that sequence by 2.



1. Find the arithmetic mean between:

a) $\frac{15}{4}$ and $\frac{19}{4}$
Solution:
AM = $\dfrac{\frac{15}{4} + \frac{19}{4}}{2}$

$= \dfrac{\frac{15+19}{4}}{2}$

$= \frac{34}{4} × \frac{1}{2}$

$= \frac{17}{4}$

c) $\frac{2}{9}$ and $\frac{6}{7}$
Solution:
AM = $\dfrac{\frac{2}{9} + \frac{6}{7}}{2}$

$= \dfrac{\frac{2×7 + 6×9}{9×7}}{2}$

$= \dfrac{\frac{14+54}{63}}{2}$

$= \frac{68}{63} × \frac{1}{2}$

$= \frac{34}{63}$


e) ($\frac{p}{2} + \frac{q}{2}$) and ($\frac{q}{2) - \frac{p}{2}$)
Solution:
AM = $\dfrac{ \frac{p}{2} + \frac{q}{2} + \frac{q}{2} - \frac{p}{2}}{2}$

$= \dfrac{2× \frac{q}{2}}{2}$

$= \frac{q}{2}$

2 a) If arithmetic mean between two numbers is 20 and first number is 28, find the second number.

Solution:
Here,
AN = 20
a = 28
Let the second number be b.
We have,
AM = $\dfrac{a + b}{2}$
$or, 20 = \dfrac{28+b}{2}$
$or, 40 = 28 + b$
$or, b = 40 - 28$
$\therefore b = 12$


3. Insert:

a) 5 arithmetic means between -7 and 17
Solution:
no of means (n) = 5
a = - 7
b = 17
Let arithmetic mean be $m_n$.

We know,
$m_n = a + nd$
And,
d = $\dfrac{b - a}{n +1}$
$= \dfrac{17 +7}{5+1}$
$= \dfrac{24}{6}$
$= 6$
For $n^{th}$ mean, $m_n = -7 + 4n$

When n = 1,
$m_1 = -7 + 4×1$
$= -7+4$
$= -3$
When n = 2,
$m_2 = -7 + 4×2$
$= -7 +8$
$= 1$
When n = 3,
$m_3 = -7 + 4×3$
$= -7 + 12$
$= 5$
When n = 4,
$m_4 = -7 + 4×4$
$= -7 + 16$
$= 9$
When n = 5,
$m_5 = -7 + 4×5$
$= -7 + 20$
$= 13$
Hence, five arithmetic means in between -7 and 17 are -3, 1, 5, 9, and 13.


c) 3 arithmetic means between 2 and 10
Solution:
Here,
no. of mean (n) = 3
a = 2
b = 10

Let arithmetic mean be $m_n$

We know,
$m_n = a + nd$
And,
d = $\dfrac{b-a}{n+1}$
$= \dfrac{10-2}{3+1}$
$= \dfrac{8}{4}$
$= 2$
For $n^{th}$ mean, $m_n = 2 + 2d$

When n = 1,
$m_1 = 2 + 2×1$
$= 4$
When n = 2,
$m_2 = 2 + 2×2$
$= 6$

When n = 3,
$m_3 = 2 +2 ×3$
$= 8$
Hence, three arithmetic mean between 2 and 10 are 4, 6 and 8.



4 a) If 13, p, q, r, 29 are in arithmetic sequence, find the values of p, q and r.

Solution:
Here,
a = 13
b = 29
no. of mean (n) = total terms - 2
= 5 - 2
= 3
We know,
Arithmetic mean ($m_n$) = $a + nd$

Also,
d = $\dfrac{b - a}{n+1}$
$= \dfrac{29-13}{3+1}$
$= \dfrac{16}{4}$
$= 4$
So, $m_n = 13 + 4n$
When n = 1,
$m_1 = p = 13 + 4×1$
$= 17$
When n = 2,
$m_2 = q = 13 + 4×2$
$= 21$
When n = 3,
$m_3 = r = 13 + 4×3$
$= 25$
Hence, the required values of p, q, and r are 17, 21, and 25.


6 a) There are 6 arithmetic means between a and b. If the second mean and last mean are 8 and 20, respectively, find a and b.

Solution:

Here,
number of arithmetic means (n) = 6

We know,
d = $\frac{b - a}{n +1}$
Given,
$m_2 = 8$ and $m_6 = 20$
We know,
$m_n = a + nd$
And,
$m_2 = a +2d$
$or, 8 = a + 2(\frac{b-a}{6+1})$

$or, 8 = a + \frac{2(b-a)}{7}$

$or, 8 = \frac{7a + 2b -2a}{7}$

$or, 8×7 = 5a + 2b$
$or, 56 = 5a + 2b$
$or, 56 - 2b = 5a$
$or, a = \frac{56-2b}{5}$ - (i)

Also,
$m_6 = a + 6d$
$or, 20 = a + 6(\frac{b -a}{6+1} )$

$or, 20 = a + \frac{6(b-a)}{7}$

$or, 20 = \frac{7a + 6b - 6a}{7}$

$or, 20 × 7 = a + 6b$
$or, 140 = a + 6b$
$or, a = 140 - 6b$ - (ii)
Solving equations (i) and (ii), we get,

$or, \frac{56-2b}{5} = 140 - 6b$

$or, 56 - 2b = 5(140 - 6b)$
$or, 56 - 2b = 700 - 30b$
$or, 30b -2b = 700 - 56$
$or, 28b = 644$
$or, b = \frac{644}{28}$
$\therefore b = 23$
Put value of b in equation (ii), we get,

$a = 140 - 6×23$
$or, a = 140 - 138$
$\therefore a = 2$
Hence, the required values of a and b are 2 and 23, respectively.


8 c) Divide 24 into three parts, which are in arithmetic sequence, such that their product is 224.

Solution:

Let the three numbers in the arithmetic sequence be a-d, a, a+d.

According to question,
$a - d + a + a + d = 24$
$or, 3a = 24$
$\therefore a = 8$

Also,
$(a - d)×a×(a +d) = 224$
$or, a (a^2 - d^2) = 224$
Put value of a now

$or, 8 (8^2 - d^2) = 224$
$or, 64 - d^2 = \frac{224}{8}$

$or, 64 - d^2 = 28$
$or, d^2 = 64 - 28$
$or, d^2 = 36$
$or, d = \sqrt{36}$
$\therefore d = \pm6$
When d = 6,
$a - d = 8 - 6 = 2$
$a = 8$
$a + d = 8+6 = 14$
When d = -6,
$a - d = 8+6 = 14$
$a = 8$
$a + d = 8 - 6 = 2$

Hence, the required parts in which the money was divided is 2,8,14 or 14,8,2.




About Readmore Optional Mathematics Book 10

Author: D. R. Simkhada
Editor: I. R. Simkhada

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Arithmetic Mean - Class 10 Solved Exercises | Readmore Optional Mathematics is a collection of the solutions related to exercises of  Arithmetic Mean chapter from Algebra Unit for Nepal's Secondary Education Examination (SEE) appearing students.

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