10 - Arithmetic Sequence and Series (Algebra)

In this post, you can check the solutions to the exercises of Arithmetic Sequence and Series 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 Arithmetic Sequence and Series.

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 term or the $n^{th}$ term of the sequence is $t_n = a + (n-1)d$.
  4. The last term of the sequence, when the number of terms is not known, is denoted by $t_n$ or b or l.


1. The first term (a) and common difference (d) of the arithmetic sequences are given below. Write down the first 4 terms.

a) a = 2 and d = 3
Solution:
We know, $t_n = a + (n-1)d$
$or, t_n = 2 + (n-1)3$
For the first four terms, a = 1,2,3,4 respectively

When n = 1
$t_1 = 2 + (1-1)3$
$\therefore t_1 = 2$
When n = 2
$t_2 = 2+(2-1)3$
$or, t_2 = 2+3$
$\therefore t_2 = 5$
When n = 3
$t_3 = 2+(3-1)3$
$or, t_3 = 2 + 6$
$\therefore t_3 = 8$
When n = 4
$t_4 = 2+(4-1)3$
$or, t_4 = 2 + 9$
$\therefore t_4 = 11$
Hence, first four terms are 2,5,8 and 11.


c) a = 20 and d = -2
Solution:
We know, $t_n = a + (n-1)d$
$or, t_n = 20 + (n-1)(-2)$
$or, t_n = 20 - (n-1)2$
For the first four terms, a = 1,2,3,4 respectively

When n = 1
$t_1 = 20 - (1-1)2$
$\therefore t_1 = 20$
When n = 2
$t_2 = 20-(2-1)2$
$or, t_2 = 20-2$
$\therefore t_2 = 18$
When n = 3
$t_3 = 20-(3-1)2$
$or, t_3 = 20-4$
$\therefore t_3 = 16$
When n = 4
$t_4 = 20-(4-1)2$
$or, t_4 = 20-6$
$\therefore t_4 = 14$
Hence, first four terms are 20,18,16 and 14.



2. Calculate the common difference and $n^{th}$ term in the following arithmetic sequence.

a) 6,9,12,15,18,21,....
Solution:
Here,
a = 6
$t_2$ = 9
Now,
Common difference (d) = $t_2 - a$

$= 9 - 6$
$= 3$
So, common difference of the sequence is 3.


And,
$n^{th} term is$
$t_n = a + (n-1)d$
$or, t_n = 6 + (n-1)3$
$or, t_n = 6 + 3n - 3$
$\therefore t_n = 3n +3$

c) 70,67,64,61,58,55,....
Solution:
Here,
a = 70
$t_2 = 67$
Now,
Common difference (d) = $t_2 - a$

$= 67 - 70$
$= -3$
So, common difference of the sequence is -3.

Now,
$n^{th} term is$
$t_n = a + (n-1)d$
$or, t_n = 70 + (n-1)(-3)$
$or, t_n = 70 - 3(n -1)$
$or, t_n = 70 -3n +3$
$\therefore t_n = 73 - 3n$

3. Calculate the first term of the arithmetic sequence whose common difference is d.

a) d = 2 and $7^{th}$ term ($t_7$) = 14
Solution:
We know,
$t_n = a + (n-1)d$
$or, t_7 = a + (7-1)2$

$or, 14 = a + 6×2$
$or, a = 14 - 12$
$\therefore a = 2$

c) d = 8 and $3^{rd}$ term ($t_3$) = 40
Solution:
We know,
$t_n = a + (n-1)d$
$or, t_3 = a + (3-1)8$
$or, 40 = a + 2×8$
$or, a = 40 - 16$
$\therefore a = 24$


4. Calculate the common difference of the arithmetic sequence whose

a) first term (a) = 21 and $3^{rd}$ term ($t_3$) = 15
Solution:
We know,
$t_n = a + (n-1)d$
$or, t_3 = 21 + (3-1)d$
$or, 15 = 21 + 2d$
$or, 2d = 15 - 21$
$or, 2d = -6$
$\therefore d = -3$

c) first term (a) = 6 and $6^{th}$ term ($t_6$) = 21
Solution:
We know,
$t_n = a + (n-1)d$
$or, t_6 = 6 + (6-1)d$
$or, 21 = 6 + 5d$
$or, 5d = 21 - 6$
$or, 5d = 15$
$\therefore d = 3$



5. How many terms are there in the following series?

a) 5+8+11+....+320
Solution:
Here,
a = 5
d = 8 - 5 = 3
$t_n$ = 320
Using formula for $n^{th}$ term, we get,
$t_n = a + (n-1)d$
$or, 320 = 5 + (n-1)3$
$or, 3(n-1) = 320-5$
$or, 3(n-1) = 315$
$or, n -1 = \frac{315}{3}$
$or, n - 1 = 105$
$\therefore n = 106$
Hence, there are 106 terms in the series.



c) 10+13+16+...+49
Solution:
Here,
a = 10
d = 13 - 10 = 3
$t_n$ = 49
Using formula for $n^{th}$ term, we get,
$t_n = a + (n-1)d$
$or, 49 = 10 + (n-2)3$
$or, 3(n-2) = 49-10$
$or, 3(n-2) = 39$
$or, (n-2) = \frac{39}{3}$

$or, n - 2 = 13$
$\therefore n = 15$
Hence, there are 15 terms in the given series.



6. Calculate the common difference and first term of the series whose:

a) 10$^{th}$ term = 23 and $32^{nd}$ term = 67
Solution:
We know,
$t_n = a + (n-1)d$
Here,
$t_10 = a + (10-1)d$
$or, 23 = a + 9d$
$or, a = 23 - 9d$ ---- (i)
And,
$t_32 = a + (32-1)d$
$or, 67 = a + 31d$
$or, a = 67 - 31d$ ---- (ii)

From equations (i) and (ii), we get,
$or, 23 - 9d = 67 - 31d$
$or, 31d - 9d = 67 - 23$
$or, 22d = 44$
$\therefore d = 2$

Now,
Put value of d in equation (i), we get,
$or, a = 23 - 9×2 = 23 - 18$

$\therefore a = 5$



c) $5^{th}$ term = 13 and $10^{th}$ term = 28
Solution:
We know,
$t_n = a + (n-1)d$
Here,
$t_5 = a + (5-1)d$
$or, 13 = a + 4d$
$or, a = 13 - 4d$ --- (i)

Also,
$t_10 = a + (n-1)d$
$or, 28 = a + (10-1)d$
$or, 28 = a + 9d$
$or, a = 28 - 9d$ --- (ii)

From equations (i) and (ii), we get,
$or, 13 -4d = 28 - 9d$
$or, 9d -4d = 28 - 13$
$or,5d = 15$
$\therefore d = 3$

Put value of d in equation (i),
$or, a = 13 - 4×3$
$or, a = 13 - 12$
$\therefore a = 1$
Hence, the required values of d and a are 3 and 1, respectively.



7 a) 6th and 17th terms of an arithmetic series are 19 and 41 respectively. Find its 100th term.

Solution:
Here,
$t_6 = 19$
$t_{17} = 41$

We have,
$t_n = a + (n-1)d$
Using formula,
$or, t_6 = 19 = a + 5d$
$or, 19 = a + 5d$
$or, a = 19 - 5d$ -- (i)
Also,
$t_{17}=  41 = a + 16d$
$or, 41 = a + 16d$
$or, a = 41 - 16d$ -- (ii)
From (i) and (ii), we get,
$or, 19 - 5d = 41 - 16d$
$or, 16d - 5d = 41 - 19$
$or, 11d = 22$
$\therefore d = 2$
Put value of d in equation (i), we get,
$or, a = 19 - 5×2 = 19-10$
$\therefore a  = 9$
Now,
$t_{100} = a + 99d$
$or, t_{100} = 9 + 99×2$
$or, t_{100} = 9 + 198$
$\therefore t_{100} = 207$

7 c) Find 20th term of an arithmetic series whose 7th and 51th terms are -3 and -355 respectively.

Solution:
Here,
$t_7 = -3$
$t_{51} = -355$
We know,
$t_n = a + (n-1)d$
Using formula,
$t_7 = -3 = a + 6d$
$or, - 3 = a + 6d$
$or, a = -3 - 6d$ -- (i)
Also,
$t_{51} = -355 = a + 50d$
$or, -355 = a + 50d$
$or, a = -355 - 50d$ -- (ii)
From equations (i) and (ii), we get,
$or, - 3 - 6d = -355 - 50d$
$or, 50d - 6d = -355 + 3$
$or, 44d = -352$
$\therefore d = -8$
Put value of d in equation (i), we get,
$or, a = -3 - 6×(-8)$
$or, a = -3 +48$
$\therefore a = 45$
Now,
$t_{20} = a + 19d$
$or, t_{20} = 45 + 19(-8)$
$or, t_{20} = 45 - 152$
$\therefore t_{20} = -107$


8 a) If the 5th term of an arithmetic series is 19 and 8th term is 31, which term is 67?

Solution:
Here,
$t_5 = 19$
$t_8 = 31$
To find: n = ? for $t_n = 67$
We know,
$t_n = a + (n-1)d$
Using formula,
$or, t_5 = a + 4d = 19$
$or, a = 18 - 4d$ - (i)
Also,
$or, t_8 = a + 7d = 31$
$or, a = 31 - 7d$ - (ii)
From equations (i) and (ii), we get,
$or, 19 - 4d = 31 - 7d$
$or, 7d - 4d = 31 - 19$

$or, 3d = 12$

$\therefore d = 4$
Put value of d in equation (i), we get,
$or, a = 19 - 4×4$
$or, a = 19 - 16$
$\therefore a = 3$
From given,
$t_n = a + (n-1)d = 67$
$or, 3 + (n-1)4 = 67$
$or, (n-1)4 = 67-3$
$or, (n-1)4 = 64$

$or, (n-1) = 16$
$\therefore n = 16+1 = 17$
Hence, the 17th term of the series is equal to 67.


9 a) If 10 times the 10th term of an AS is equal to 15times the 15th term, find the common difference if first term of the series is 48.

Solution:
Given,
$a = 48$
$10 × t_{10} = 15 × t_{15}$
$or, 10(a + 9d) = 15(a + 14d)$

$or, 2(a + 9d) = 3(a + 14d)$
$or, 2(48 + 9d) = 3(48 + 14d)$

$or, 2×48 + 18d = 3×48 + 42d$
$or, 2×48 - 3×48 = 42d - 18d$
$or, 48(2-3) = 24d$
$or, -48 = 24d$
$\therefore d = -2$


10 a) Divide 45 into three parts which are in AS such that their product is 1875.

10 c) The sum of 3 numbers in an AS is 12 and the sum of their squares is 66. Find the numbers. 

10 e) The sum of 3 numbers in an AS is 12 and sum of their cubes is 288. Find the numbers.

Solution:
Let the three numbers in an AS be (a+d), a and (a-d).

Condition I,
Sum of 3 numbers is 12
$or, (a +d) + a + (a -d) = 12$

$or, 3a = 12$
$\therefore a = 4$
Condition II,
Sum of cubes of 3 numbers is 288
$or, (a +d)^3 + a^3 +(a -d)^3 = 288$

$or, (4+d)^3 + 4^3 + (4-d)^3 = 288$

$or, (4+d)^3 + (4-d)^3 = 288 - 4^3$

[Using formula (a + b)³ + (a - b)³ = 2a³ + 6ab²]

$or, 2×(4)^3 + 6×4×d^2 = 224$
$or, 2×64 + 24d^2 = 224$
$or, 24d^2 = 224- 128$
$or, 24d^2 = 96$
$or, d^2 = 4$

$\therefore d = \pm 2$
Now,
Required numbers are:
When d = +2,
$(a - d) = (4-2) = 2$
$a = 4$
$(a +d) = (4+2) = 6$
When d = -2,
$(a - d) = 4-(-2) = 6$
$a = 4$
$(a +d) = 4+(-2) = 2$
Hence, the required numbers are either 2,4,6 or 6,4,2.



11 a) The 10th, 81st and last terms of an AS are 95, 450 and 1050 respectively. Find the number of terms in the series.

Solution:
Given,
$t_{10} = 95$
$t_{81} = 450$
$t_n = 1050$
We have,
$t_n = a + (n-1)d$
Using formula,
$t_{10} = a + 9d = 95$
$or, a = 95 - 9d$ - (i)
Also,
$t_{81} = a + 80d = 450$
$or, a = 450 - 80d$ - (ii)
From equations (i) and (ii), we get,
$or, 95 - 9d = 450 - 80d$
$or, 80d - 9d = 450 - 95$
$or, 71d = 355$
$or, d = \dfrac{355}{71}$
$\therefore d = 5$
Put value of d in equation (i), we get,
$or, a = 95 - 9×5 = 95 - 45$
$\therefore a = 50$
Now,
$t_n = a + (n-1)d = 1050$
$or, 50 + (n-1)5 = 1050$
$or, 5(n-1) = 1050 - 50$
$or, 5(n-1) = 1000$
$or, (n-1) = 200$
$\therefore n = 200+1 = 201$
Hence, there are 201 terms in the given arithmetic sequence.

12 a) The nth term of the series 16+14+12+... And 22+19+16+... are same. Find n.

Solution:
First series: 16+14+12+....
$a_1= 16$
$d_1= 14-16 = -2$
Second series: 22+19+16
$a_2= 22$
$d_2 = 19-22 = -3$
Given,
$t_{n1} = t_{n2{$
$or, a_1 + (n-1)d_1 = a_2 + (n-1)d_2$

$or, 16 + (n-1)(-2) = 22 + (n-1)(-3)$

$or, 16 - 2n + 2 = 22 -3n +3$
$or, 3n - 2n = 25 - 18$
$\therefore n = 7$
Hence, the required value of n is 7.

About Readmore Optional Mathematics Book 10

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

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

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