# Grade 10 Optional Mathematics: Sequence and Series Study Guide
## Unit: Algebra | Chapter: Sequence and Series | Complete Solutions

---

## 🎯 Chapter Introduction

Welcome to the comprehensive study guide for the **Sequence and Series** chapter. This note is designed to make complex concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to a wide range of problems. By the end of this guide, you will have mastered Arithmetic and Geometric progressions. 📈

### Key Learning Areas
1.  **Definitions:** Understanding Sequences, Series, and their different types.
2.  **Arithmetic Progression (A.P.):** Working with sequences that have a common difference.
3.  **Geometric Progression (G.P.):** Working with sequences that have a common ratio.
4.  **Means:** Calculating and inserting Arithmetic and Geometric Means.
5.  **Summation:** Finding the sum of a given number of terms in a series.

---

## 🔑 Key Concepts & Formulas

Before solving the problems, let's review the fundamental concepts and formulas.

### 1. Core Definitions

* **Sequence:** A set of numbers arranged in a definite order according to a specific mathematical rule.
    * *Example:* `1, 3, 5, 7, ...` is a sequence where each term is 2 more than the previous one.

* **Series:** The sum of the terms in a sequence.
    * *Example:* `1 + 3 + 5 + 7 + ...` is the series corresponding to the sequence above.

* **Arithmetic Progression (A.P.):** A sequence where the difference between any two consecutive terms is constant. This constant is called the **common difference (d)**.
    * *Example:* `5, 9, 13, 17, ...` (common difference is 4).

* **Geometric Progression (G.P.):** A sequence where the ratio of any term to its preceding term is constant. This constant is called the **common ratio (r)**.
    * *Example:* `3, 6, 12, 24, ...` (common ratio is 2).

* **Arithmetic Mean (A.M.):** A term (or terms) inserted between two given terms that maintains the arithmetic progression.

* **Geometric Mean (G.M.):** A term (or terms) inserted between two given terms that maintains the geometric progression.

---

### 2. Formulas for Arithmetic Progression (A.P.)

Let **a** be the first term, **d** be the common difference, and **n** be the number of terms.

* **nth Term ($t_n$):**
    $$t_n = a + (n-1)d$$

* **Sum of First n Terms ($S_n$):**
    $$S_n = \frac{n}{2}[2a + (n-1)d]$$

* **Arithmetic Mean between a and b:**
    $$A.M. = \frac{a+b}{2}$$

* **Inserting 'n' Arithmetic Means between a and b:**
    The common difference `d` is found using:
    $$d = \frac{b-a}{n+1}$$

---

### 3. Formulas for Geometric Progression (G.P.)

Let **a** be the first term, **r** be the common ratio, and **n** be the number of terms.

* **nth Term ($t_n$):**
    $$t_n = ar^{n-1}$$

* **Sum of First n Terms ($S_n$):**
    $$S_n = \frac{a(r^n - 1)}{r-1} \quad (\text{if } r > 1)$$
    $$S_n = \frac{a(1 - r^n)}{1-r} \quad (\text{if } r < 1)$$

* **Geometric Mean between a and b:**
    $$G.M. = \sqrt{ab}$$

* **Inserting 'n' Geometric Means between a and b:**
    The common ratio `r` is found using:
    $$r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$$

---

## 📚 Exercise Solutions

### Short Questions

***

#### Question 1: Arithmetic and Geometric Mean
**Find the arithmetic mean and geometric mean between 4 and 16.**

**Solution:**
Here, $a = 4$ and $b = 16$.

**Arithmetic Mean (A.M.):**
$$
\begin{aligned}
A.M. &= \frac{a+b}{2} \\
&= \frac{4+16}{2} \\
&= \frac{20}{2} = 10
\end{aligned}
$$
**Geometric Mean (G.M.):**
$$
\begin{aligned}
G.M. &= \sqrt{ab} \\
&= \sqrt{4 \times 16} \\
&= \sqrt{64} = 8
\end{aligned}
$$
The A.M. is **10** and the G.M. is **8**.

***

#### Question 2: Arithmetic and Geometric Mean
**Find the arithmetic mean and geometric mean between 3 and 27.**

**Solution:**
Here, $a = 3$ and $b = 27$.

**Arithmetic Mean (A.M.):**
$$
A.M. = \frac{3+27}{2} = \frac{30}{2} = 15
$$
**Geometric Mean (G.M.):**
$$
G.M. = \sqrt{3 \times 27} = \sqrt{81} = 9
$$
The A.M. is **15** and the G.M. is **9**.

***

#### Question 3: Geometric Sequence
**If 4, a, and 16 are in a geometric sequence, find the value of a.**

**Solution:**
In a geometric sequence, the middle term is the geometric mean of the first and last terms.
$$
\begin{aligned}
a &= \sqrt{4 \times 16} \\
a &= \sqrt{64} \\
a &= 8
\end{aligned}
$$
Alternatively, the ratio of consecutive terms is constant:
$$
\begin{aligned}
\frac{a}{4} &= \frac{16}{a} \\
a^2 &= 64 \\
a &= 8
\end{aligned}
$$
Thus, **$a=8$**.

***

#### Question 4: Relation between Means
**If the arithmetic mean between 4 and x is 34, find their geometric mean.**

**Solution:**
First, use the A.M. formula to find $x$.
$$
\begin{aligned}
A.M. &= \frac{4+x}{2} \\
34 &= \frac{4+x}{2} \\
68 &= 4+x \\
x &= 64
\end{aligned}
$$
Now, find the G.M. between 4 and 64.
$$
G.M. = \sqrt{4 \times 64} = \sqrt{256} = 16
$$
The geometric mean is **16**.

***

#### Question 5: Relation between Means
**If the geometric mean of 2 and x is 4, find their arithmetic mean.**

**Solution:**
First, use the G.M. formula to find $x$.
$$
\begin{aligned}
G.M. &= \sqrt{2x} \\
4 &= \sqrt{2x} \\
16 &= 2x \\
x &= 8
\end{aligned}
$$
Now, find the A.M. between 2 and 8.
$$
A.M. = \frac{2+8}{2} = \frac{10}{2} = 5
$$
The arithmetic mean is **5**.

***

#### Question 6: Geometric Sequence
**If $m + 2$, $m + 8$, and $17 + m$ are in geometric sequence, find the value of m.**

**Solution:**
Since the terms are in a G.P., the ratio of consecutive terms is constant.
$$
\begin{aligned}
\frac{m+8}{m+2} &= \frac{17+m}{m+8} \\
(m+8)^2 &= (m+2)(17+m) \\
m^2 + 16m + 64 &= 17m + m^2 + 34 + 2m \\
m^2 + 16m + 64 &= m^2 + 19m + 34 \\
16m + 64 &= 19m + 34 \\
30 &= 3m \\
m &= 10
\end{aligned}
$$
Thus, **$m=10$**.

***

#### Question 7: Finding the Term Number in an A.P.
**Which term of the series $5 + 9 + 13 + ...$ is 85?**

**Solution:**
This is an arithmetic series with:
First term, $a = 5$.
Common difference, $d = 9 - 5 = 4$.
Let the nth term, $t_n = 85$.

Using the formula $t_n = a + (n-1)d$:
$$
\begin{aligned}
85 &= 5 + (n-1)4 \\
80 &= (n-1)4 \\
20 &= n-1 \\
n &= 21
\end{aligned}
$$
Thus, 85 is the **21st term**.

***

#### Question 8: Finding the Term Number in a G.P.
**Which term of the series $3 + 6 + 12 + ...$ is 192?**

**Solution:**
This is a geometric series with:
First term, $a = 3$.
Common ratio, $r = 6/3 = 2$.
Let the nth term, $t_n = 192$.

Using the formula $t_n = ar^{n-1}$:
$$
\begin{aligned}
192 &= 3 \cdot (2)^{n-1} \\
64 &= (2)^{n-1} \\
2^6 &= 2^{n-1} \\
6 &= n-1 \\
n &= 7
\end{aligned}
$$
Thus, 192 is the **7th term**.

***

#### Question 9: Sum of a G.P.
**Find the sum of the series $9 + 3 + 1 + ... + \frac{1}{27}$.**

**Solution:**
This is a geometric series with $a=9$ and $r = 3/9 = 1/3$. The last term is $t_n = 1/27$.

**Step 1: Find the number of terms (n).**
$$
\begin{aligned}
t_n &= ar^{n-1} \\
\frac{1}{27} &= 9 \cdot \left(\frac{1}{3}\right)^{n-1} \\
\frac{1}{243} &= \left(\frac{1}{3}\right)^{n-1} \\
\left(\frac{1}{3}\right)^5 &= \left(\frac{1}{3}\right)^{n-1} \\
5 &= n-1 \\
n &= 6
\end{aligned}
$$
**Step 2: Find the sum of the 6 terms ($S_6$).**
Since $r < 1$, we use $S_n = \frac{a(1-r^n)}{1-r}$.
$$
\begin{aligned}
S_6 &= \frac{9(1 - (1/3)^6)}{1 - 1/3} \\
&= \frac{9(1 - 1/729)}{2/3} \\
&= \frac{9(728/729)}{2/3} \\
&= 9 \times \frac{728}{729} \times \frac{3}{2} \\
&= \frac{364}{27}
\end{aligned}
$$
The sum of the series is **$\frac{364}{27}$**.

***

#### Question 10: Product of Terms in a G.P.
**If the third term of a geometric series is 3, find the product of the first five terms.**

**Solution:**
Let the first five terms of the G.P. be $a, ar, ar^2, ar^3, ar^4$.
We are given that the third term, $t_3 = ar^2 = 3$.

The product of the first five terms is:
$$
\begin{aligned}
P_5 &= (a)(ar)(ar^2)(ar^3)(ar^4) \\
&= a^5 r^{0+1+2+3+4} \\
&= a^5 r^{10} \\
&= (ar^2)^5
\end{aligned}
$$
Since we know $ar^2 = 3$:
$$
P_5 = (3)^5 = 243
$$
The product is **243**.

***

#### Question 11: Finding Terms in a G.P.
**If $\frac{1}{3}, p, q, 9$ are in a geometric sequence, find the values of p and q.**

**Solution:**
Here, $a = 1/3$ and the 4th term, $t_4 = 9$.
First, find the common ratio $r$.
$$
\begin{aligned}
t_4 &= ar^{4-1} \\
9 &= \left(\frac{1}{3}\right)r^3 \\
27 &= r^3 \\
r &= 3
\end{aligned}
$$
Now find $p$ and $q$:
$p$ is the 2nd term: $p = ar = (\frac{1}{3})(3) = 1$.
$q$ is the 3rd term: $q = ar^2 = (\frac{1}{3})(3)^2 = 3$.

Thus, **$p=1$ and $q=3$**.

***

#### Question 12: Finding Common Ratio
**Find the common ratio of a geometric sequence, whose first term is 2 and the third term is 242.**

**Solution:**
Given $a=2$ and $t_3 = 242$.
$$
\begin{aligned}
t_3 &= ar^{3-1} \\
242 &= 2 \cdot r^2 \\
121 &= r^2 \\
r &= \pm 11
\end{aligned}
$$
The common ratio is **$\pm 11$**.

***

#### Question 13: Finding Terms in an A.P.
**If $5, x, y, 11$ is an arithmetic series. Find the value of x and y.**

**Solution:**
Here, $a = 5$ and the 4th term, $t_4 = 11$.
First, find the common difference $d$.
$$
\begin{aligned}
t_4 &= a + (4-1)d \\
11 &= 5 + 3d \\
6 &= 3d \\
d &= 2
\end{aligned}
$$
Now find $x$ and $y$:
$x$ is the 2nd term: $x = a+d = 5+2 = 7$.
$y$ is the 3rd term: $y = a+2d = 5+2(2) = 9$.

Thus, **$x=7$ and $y=9$**.

***

#### Question 14: Sum of a G.P.
**Find the sum of the series: 2 + 4 + 8 + ... upto 8 terms.**

**Solution:**
This is a G.P. with $a=2$, $r=2$, and $n=8$.
Since $r > 1$, we use the formula $S_n = \frac{a(r^n - 1)}{r-1}$.
$$
\begin{aligned}
S_8 &= \frac{2(2^8 - 1)}{2-1} \\
&= \frac{2(256 - 1)}{1} \\
&= 2(255) \\
&= 510
\end{aligned}
$$
The sum of the first 8 terms is **510**.

***

### Long Questions

***

#### Question 15: A.P. and G.P. Relationship
**The product of the first five terms of a geometric series is 243. If the third term of the geometric series is equal to the tenth term of an arithmetic series, find the sum of the first 19 terms of the arithmetic series.**

**Solution:**
**For the G.P.:**
Product $P_5 = (a)(ar)(ar^2)(ar^3)(ar^4) = a^5r^{10} = (ar^2)^5$.
$$
\begin{aligned}
(ar^2)^5 &= 243 \\
(ar^2)^5 &= 3^5 \\
ar^2 &= 3
\end{aligned}
$$
So, the third term of the G.P. ($t_3$) is 3.

**For the A.P.:**
We are given that the tenth term ($t_{10}$) of the A.P. is equal to the third term of the G.P.
$$
t_{10} = a + (10-1)d = a + 9d = 3
$$
**Sum of the A.P.:**
We need to find the sum of the first 19 terms ($S_{19}$).
$$
\begin{aligned}
S_{19} &= \frac{19}{2}[2a + (19-1)d] \\
&= \frac{19}{2}[2a + 18d] \\
&= \frac{19}{2} \cdot 2(a + 9d) \\
&= 19(a + 9d)
\end{aligned}
$$
Since we know $a+9d = 3$:
$$
S_{19} = 19(3) = 57
$$
The sum of the first 19 terms is **57**.

***

#### Question 16: Finding Terms from Means
**There are five geometric means between a and b. If the second mean is 2 and the last mean is 16, find the values of a and b.**

**Solution:**
The sequence is $a, m_1, m_2, m_3, m_4, m_5, b$.
This is a G.P. with $n = 5+2 = 7$ terms.
$a$ is the 1st term.
$m_2$ (second mean) is the 3rd term of the sequence: $t_3 = ar^2 = 2$.
$m_5$ (last mean) is the 6th term of the sequence: $t_6 = ar^5 = 16$.

Divide the two equations:
$$
\begin{aligned}
\frac{ar^5}{ar^2} &= \frac{16}{2} \\
r^3 &= 8 \\
r &= 2
\end{aligned}
$$
Now find $a$ using $ar^2 = 2$:
$$
\begin{aligned}
a(2)^2 &= 2 \\
4a &= 2 \\
a &= \frac{1}{2}
\end{aligned}
$$
Find $b$, which is the 7th term ($t_7$):
$$
b = ar^6 = \left(\frac{1}{2}\right)(2)^6 = \frac{64}{2} = 32
$$
Thus, **$a = 1/2$ and $b = 32$**.

***

#### Question 17: Sum of an A.P. from Sums
**If the sum of the first 9 terms of an arithmetic series is 72 and the sum of the first 17 terms is 289, find the sum of the first 25 terms.**

**Solution:**
Given $S_9 = 72$ and $S_{17} = 289$.
For $S_9$:
$$
\begin{aligned}
\frac{9}{2}(2a + 8d) &= 72 \\
9(a+4d) &= 72 \\
a+4d &= 8 \implies a = 8-4d \quad \text{(Eq. 1)}
\end{aligned}
$$
For $S_{17}$:
$$
\begin{aligned}
\frac{17}{2}(2a + 16d) &= 289 \\
17(a+8d) &= 289 \\
a+8d &= 17 \quad \text{(Eq. 2)}
\end{aligned}
$$
Substitute Eq. 1 into Eq. 2:
$$
\begin{aligned}
(8-4d)+8d &= 17 \\
8+4d &= 17 \\
4d &= 9 \implies d=9/4
\end{aligned}
$$
Now find $a$: $a = 8 - 4(9/4) = 8-9 = -1$.

Find $S_{25}$:
$$
\begin{aligned}
S_{25} &= \frac{25}{2}[2(-1) + (25-1)(9/4)] \\
&= \frac{25}{2}[-2 + 24(9/4)] \\
&= \frac{25}{2}[-2 + 6 \times 9] \\
&= \frac{25}{2}[-2 + 54] = \frac{25}{2}(52) \\
&= 25 \times 26 = 650
\end{aligned}
$$
The sum of the first 25 terms is **650**.

***

#### Question 18: Sum of a G.P. from Two Terms
**If the second term and fifth term of the geometric series are 15 and 405 respectively, find the sum of the first 6 terms of the series.**

**Solution:**
Given $t_2 = ar = 15$ and $t_5 = ar^4 = 405$.
Divide the two equations:
$$
\begin{aligned}
\frac{ar^4}{ar} &= \frac{405}{15} \\
r^3 &= 27 \\
r &= 3
\end{aligned}
$$
Find $a$: $a(3) = 15 \implies a=5$.
Find $S_6$ (since $r>1$):
$$
\begin{aligned}
S_6 &= \frac{5(3^6-1)}{3-1} \\
&= \frac{5(729-1)}{2} \\
&= \frac{5(728)}{2} = 5(364) = 1820
\end{aligned}
$$
The sum of the first 6 terms is **1820**.

***

#### Question 19: Sum of a G.P. from Two Terms
**Find the sum of the first 9 terms of a geometric series whose third term and seventh term are 20 and 320 respectively.**

**Solution:**
Given $t_3 = ar^2 = 20$ and $t_7 = ar^6 = 320$.
Divide the two equations:
$$
\begin{aligned}
\frac{ar^6}{ar^2} &= \frac{320}{20} \\
r^4 &= 16 \\
r &= 2
\end{aligned}
$$
Find $a$: $a(2)^2 = 20 \implies 4a=20 \implies a=5$.
Find $S_9$:
$$
\begin{aligned}
S_9 &= \frac{5(2^9-1)}{2-1} \\
&= 5(512-1) \\
&= 5(511) = 2555
\end{aligned}
$$
The sum of the first 9 terms is **2555**.

***

#### Question 20: Inserting Arithmetic Means
**Insert four arithmetic means between 5 and 25.**

**Solution:**
We are inserting $n=4$ means between $a=5$ and $b=25$.
First, find the common difference $d$:
$$
d = \frac{b-a}{n+1} = \frac{25-5}{4+1} = \frac{20}{5} = 4
$$
The means are:
* $m_1 = a+d = 5+4 = 9$
* $m_2 = a+2d = 5+2(4) = 13$
* $m_3 = a+3d = 5+3(4) = 17$
* $m_4 = a+4d = 5+4(4) = 21$

The four arithmetic means are **9, 13, 17, 21**.

***

#### Question 21: Sum of a G.P.
**Find the sum of the first 8 terms of a geometric series whose fourth term is 54 and common ratio is 3.**

**Solution:**
Given $t_4 = 54$ and $r=3$.
First, find the first term $a$.
$$
\begin{aligned}
t_4 &= ar^{3} \\
54 &= a(3)^3 \\
54 &= 27a \\
a &= 2
\end{aligned}
$$
Now, find the sum of the first 8 terms ($S_8$):
$$
\begin{aligned}
S_8 &= \frac{a(r^8-1)}{r-1} \\
&= \frac{2(3^8-1)}{3-1} \\
&= \frac{2(6561-1)}{2} \\
&= 6560
\end{aligned}
$$
The sum is **6560**.

***

#### Question 22: Sum and Product of an A.P.
**The sum of three numbers in an arithmetic progression is 51 and the sum of their squares is 939. Find the numbers.**

**Solution:**
Let the three terms be $a-d, a, a+d$.
From the sum:
$$
\begin{aligned}
(a-d) + a + (a+d) &= 51 \\
3a &= 51 \\
a &= 17
\end{aligned}
$$
From the sum of squares:
$$
\begin{aligned}
(a-d)^2 + a^2 + (a+d)^2 &= 939 \\
(17-d)^2 + 17^2 + (17+d)^2 &= 939 \\
(289 - 34d + d^2) + 289 + (289 + 34d + d^2) &= 939 \\
867 + 2d^2 &= 939 \\
2d^2 &= 72 \\
d^2 &= 36 \\
d &= \pm 6
\end{aligned}
$$
If $d=6$, the numbers are $11, 17, 23$.
If $d=-6$, the numbers are $23, 17, 11$.
The numbers are **11, 17, and 23**.

***

#### Question 23: Sum and Product of an A.P.
**Three terms in an arithmetic progression have sum 21 and product 315. Find the terms.**

**Solution:**
Let the three terms be $a-d, a, a+d$.
From the sum:
$$
(a-d) + a + (a+d) = 21 \implies 3a = 21 \implies a=7
$$
From the product:
$$
\begin{aligned}
(a-d)(a)(a+d) &= 315 \\
(7-d)(7)(7+d) &= 315 \\
(49-d^2)(7) &= 315 \\
49-d^2 &= 45 \\
d^2 &= 4 \\
d &= \pm 2
\end{aligned}
$$
If $d=2$, the numbers are $5, 7, 9$.
If $d=-2$, the numbers are $9, 7, 5$.
The terms are **5, 7, and 9**.

***

#### Question 24: Sum of an A.P. from Two Terms
**The ninth and nineteenth terms of an arithmetic series are 40 and 60 respectively. Find the sum of the first 25 terms of the series.**

**Solution:**
Given: $t_9 = a+8d = 40$ and $t_{19} = a+18d = 60$.
Subtract the first equation from the second:
$$
(a+18d) - (a+8d) = 60-40 \implies 10d=20 \implies d=2
$$
Find $a$: $a+8(2)=40 \implies a+16=40 \implies a=24$.
Find $S_{25}$:
$$
\begin{aligned}
S_{25} &= \frac{25}{2}[2(24) + (25-1)2] \\
&= \frac{25}{2}[48 + 24(2)] \\
&= \frac{25}{2}[48+48] = \frac{25}{2}(96) \\
&= 25(48) = 1200
\end{aligned}
$$
The sum is **1200**.

***

#### Question 25: A.P. Word Problem
**In an arithmetic sequence, the sixth term is equal to three times the fourth term and the sum of the first three terms is -12. Find the sum of the first ten terms.**

**Solution:**
From the first condition: $t_6 = 3 \times t_4$.
$$
\begin{aligned}
a+5d &= 3(a+3d) \\
a+5d &= 3a+9d \\
-2a &= 4d \\
a &= -2d \quad \text{(Eq. 1)}
\end{aligned}
$$
From the second condition, let the first three terms be $a-d, a, a+d$.
Their sum is $(a-d)+a+(a+d) = 3a = -12 \implies a=-4$.

Substitute $a=-4$ into Eq. 1:
$-4 = -2d \implies d=2$.

Find $S_{10}$:
$$
\begin{aligned}
S_{10} &= \frac{10}{2}[2(-4) + (10-1)2] \\
&= 5[-8 + 9(2)] \\
&= 5[-8+18] = 5(10) = 50
\end{aligned}
$$
The sum is **50**.

***

#### Question 26: A.P. Word Problem
**The sum of the first eight terms of an arithmetic series is 180 and its fifth term is five times the first term. Find the first term and common difference.**

**Solution:**
From the first condition: $t_5 = 5a$.
$$
a+4d = 5a \implies 4d=4a \implies d=a \quad \text{(Eq. 1)}
$$
From the second condition: $S_8 = 180$.
$$
\begin{aligned}
\frac{8}{2}[2a+(8-1)d] &= 180 \\
4[2a+7d] &= 180 \\
2a+7d &= 45
\end{aligned}
$$
Substitute $d=a$ into this equation:
$$
2a+7a = 45 \implies 9a = 45 \implies a=5
$$
Since $d=a$, then $d=5$.
The first term is **5** and the common difference is **5**.

***

#### Question 27: Sum of a G.P. from Sums
**In a geometric series, the sum of the first three terms is 1 and the sum of its first six terms is 28. Find the common ratio and first term.**

**Solution:**
Given $S_3 = \frac{a(r^3-1)}{r-1} = 1$ and $S_6 = \frac{a(r^6-1)}{r-1} = 28$.
We can rewrite $S_6$:
$$
S_6 = \frac{a(r^3-1)(r^3+1)}{r-1} = \left[\frac{a(r^3-1)}{r-1}\right](r^3+1) = S_3(r^3+1)
$$
Now substitute the known values:
$$
\begin{aligned}
28 &= 1(r^3+1) \\
27 &= r^3 \\
r &= 3
\end{aligned}
$$
Now find $a$ using the equation for $S_3$:
$$
\begin{aligned}
\frac{a(3^3-1)}{3-1} &= 1 \\
\frac{a(26)}{2} &= 1 \\
13a &= 1 \\
a &= \frac{1}{13}
\end{aligned}
$$
The common ratio is **3** and the first term is **1/13**.