Sequence and Series Miscellaneous

Sequences and series are fundamental concepts in mathematics with applications ranging from basic arithmetic to advanced calculus. A sequence is an ordered list of numbers, while a series is the sum of the terms of a sequence. In this article, we will explore various types of miscellaneous sequences and series, including natural numbers, squares and cubes of natural numbers, arithmetic-geometric progressions (AGP), and techniques for inserting arithmetic and geometric progressions between two numbers. For better understanding, we will also discuss some solved examples.

The sequence of the first n natural numbers

A sequence of natural numbers is simply an ordered list of the first $n$ natural numbers i.e. 1, 2, 3, 4, ……, $n$.

The sum of the first $n$ natural number is given by the formula

$S_n = \frac{n(n+1)}{2}$

For example, the sum of the first 10 natural numbers is $\frac{10(10+1)}{2}$ = 55.

The sequence of squares of the first n natural numbers

A sequence of squares of n natural numbers consists of numbers that are the squares of the first $n$ natural numbers, i.e. $1^2, 2^2, 3^2, 4^2, ……., n^2$.

The sum of the squares of the first $n$ natural numbers is given by the formula,

$S_n = \frac{n(n+1)(2n+1)}{6}$

For example, the sum of the squares of the first 5 natural numbers is

$\frac{5(5+1)(2×5 +1)}{6}$ = 55

The sequence of cubes of the first n natural numbers

A sequence of cubes of the first n natural numbers consists of numbers that are the cubes of the first $n$ natural numbers, i.e. $1^3, 2^3, 3^3, 4^3, ……., n^3$.

The sum of the cubes of the first $n$ natural numbers is given by the formula,

$S_n = [\frac{n(n+1)}{2}]^2$

For example, the sum of the cubes of the first 5 natural numbers is

$[\frac{5(5+1)}{2}]^2$ = 225

Arithmetic- Geometric Progression (AGP)

An arithmetic-geometric progression (AGP) is a sequence where each term is the product of the terms of an arithmetic progression (AP) and a geometric progression (GP).

The general form of an AGP is $a, (a+d)r, (a+2d)r^2, (a+3d)r^3, …….$

where $a$ is the initial term, $d$ is the common difference, and $r$ is the common ratio.

The nth term of an AGP is obtained by multiplying the corresponding term of the arithmetic progression (AP) and the geometric progression (GP).

So, the nth term is given by the formula $t_n = [a+(n-1)d]r^{n-1}$.

The sum of the first $n$ terms of an AGP is given by the formula

$S_n = \displaystyle\sum_{k=1}^{n} [a+(k-1)d]r^{k-1}$, which can be further simplified as

$S_n = \frac{a - [a+(n-1)d]r^n}{1-r} + \frac{dr(1-r^{n-1})}{(1-r)^2}$

The sum to infinity is given by $S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$.

Inserting an AP between two numbers

Let the two numbers be $a$ and $b$. We will need to insert $n$ terms of AP between them.

Let the $n$ terms be $a_1, a_2, a_3, ……., a_n$.

So, the sequence will be $a, a_1, a_2, a_3, ……., a_n, b$ with a total of $(n + 2)$ terms and let the common difference be $d$.

Where $(n + 2)^{th}$ term is $b$ i.e. $a + (n + 2 - 1)d = b$

⇒ $d = \frac{b-a}{n+1}$

Now, the terms of the sequence become,

$a_1 = a + d$

$a_2 = a + 2d$

$a_3 = a + 3d$

.

.

.

$a_n = a + nd$

For example, let’s insert 5 terms of AP between 3 and 21.

⇒ Here $a$ = 3, $b$ = 21 and $n$ = 5

So, the common difference ($d$) = $\frac{b-a}{n+1}$ = $\frac{21-3}{5+1}$ = 3

Now, the terms are (3+3), (3+2×3), (3+3×3), (3+4×3), (3+5×3) i.e. 6, 9, 12, 15, 18.

So, the final sequence is 3, 6, 9, 12, 15, 18, and 21.

Inserting a GP between two numbers

Let the two numbers be $x$ and $y$. We will need to insert $n$ terms of GP between them.

Let the $n$ terms be $a_1, a_2, a_3, ……., a_n$.

So, the sequence will be $x, a_1, a_2, a_3, ……., a_n, y$ with a total of $(n + 2)$ terms and let the common ratio be $r$.

Where $(n + 2)^{th}$ term is $y$ i.e. $xr^{n+2-1} = y$

⇒ $r = (\frac{y}{x})^{\frac{1}{n+1}}$

Now, the terms of the sequence become,

$a_1 = ar$

$a_2 = ar^2$

$a_3 = ar^3$

.

.

.

$a_n = ar^n$

For example, let’s insert 4 terms of GP between 5 and 160.

⇒ Here x = 5, y = 160 and n = 4

Thus, the common ratio (r) = $(\frac{y}{x})^{\frac{1}{n+1}}$ = $(\frac{160}{5})^{\frac{1}{4+1}}$

So, $r = 32^{\frac{1}{5}} = 2$

Now, the terms are $(5×2), (5×2^2), (5×2^3), (5×2^4)$ i.e. 10, 20, 40, 80.

So, the final sequence is 5, 10, 20, 40, 80, and 160.

Deduction of a sequence if a general term is given

If the general term $t_n$ of a sequence is given, then the sequence can be deduced by substituting consecutive integer values of $n$.

For example, if the general term of a sequence is $t_n=2n+1$,

Then $t_1=2(1)+1=3$

$t_1=2(1)+1=3$

$t_1=2(2)+1=5$

$t_1=2(3)+1=7$

$t_1=2(4)+1=9$

.

.

.

So, the sequence is 3, 5, 7, 9, …….

Solved Examples

Q.1.

Show that the sum of $(m + n)^{th}$ and $(m – n)^{th}$ terms of an AP is equal to twice the $m$th term.

Solution:

Let the first term and the common difference of the AP be $a$ and $d$.

Now, the $(m+n)^{th}$ term is $t_{m+n} = a+(m+n-1)d$

And the $(m-n)^{th}$ term is $t_{m-n} = a+(m-n-1)d$

So, the sum of these two terms = $t_{m+n}+t_{m-n}=a+(m+n-1)d+a+(m-n-1)d=2[a+(m-1)d]$

Which is twice the $m^{th}$ term.

Hence, the sum of $(m + n)^{th}$ and $(m – n)^{th}$ terms of an AP is equal to twice the $m^{th}$ term.

Q.2.

The sum of the first 20 term of the series $\frac{1}{5×6}+\frac{1}{6×7}+\frac{1}{7×8}+....$ is:

1. 0.16

2. 1.6

3. 16

4. 0.016

Hint: Break the all fractions term as $\frac{1}{5 \times 6}=\frac{1}{5}-\frac{1}{6}$ and add all terms.

Solution:

Given: The sum of the first 20 terms of the series $\frac{1}{5×6}+\frac{1}{6×7}+\frac{1}{7×8}+....$

So, $a_{1} =5$ and $a_{20} =5+(20-1)×1=24$

20th term $=\frac{1}{24×25}$

$=\frac{1}{5×6}+\frac{1}{6×7}+\frac{1}{7×8}+.... + \frac{1}{24×25}$

$=\frac{1}{5}-\frac{1}{6} + \frac {1}{6} - \frac{1}{7}+\frac{1}{7}-\frac{1}{8}+..............+\frac{1}{24}-\frac{1}{25}$

$=\frac{1}{5}-\frac{1}{25}$

$=\frac{4}{25}$

$= 0.16$

Hence, the correct answer is option (1).

Q.3.

What is the value of S $=\frac{1}{1×3×5}+\frac{1}{1×4}+\frac{1}{3×5×7}+\frac{1}{4×7}+\frac{1}{5×7×9}+\frac{1}{7×10}+....$ Up to 20 terms.

1. $\frac{6179}{15275}$

2. $\frac{6070}{14973}$

3. $\frac{7191}{15174}$

4. $\frac{5183}{16423}$

Hint: Arrange the same type of terms on one side and other types on one side, then take $\frac{1}{4}$ common from the first type and $\frac{1}{3}$ common from 2nd type.

Solution:

Given:

$S=\frac{1}{1×3×5}+\frac{1}{1×4}+\frac{1}{3×5×7}+\frac{1}{4×7}+\frac{1}{5×7×9}+\frac{1}{7×10}+....$ Up to 20 terms.

Rewriting it like,

$\frac{1}{1×3×5}+\frac{1}{3×5×7}+\frac{1}{5×7×9}+...+\frac{1}{1×4}+\frac{1}{4×7}+\frac{1}{7×10}+...$Up to 20 terms.

$=\frac{1}{4}[\frac{1}{1×3}-\frac{1}{3×5}+\frac{1}{3×5}-\frac{1}{5×7}+\frac{1}{5×7}-\frac{1}{7×9}+...–\frac{1}{21×23}]+\frac{1}{3}[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...-\frac{1}{31}]$

It will cancel out most of the terms except the first and last term.

$=\frac{1}{4}[\frac{1}{1×3}-\frac{1}{21×23}]+\frac{1}{3}[1-\frac{1}{31}]$

$=\frac{40}{483}+\frac{10}{31}$

$=\frac{6070}{14973}$

Hence, the correct answer is option (2).

Q.4.

What is the value of $14^{3}+16^{3}+18^{3}+...+30^{3}$?

1. 134576

2. 120212

3. 115624

4. 111672

Hint: Use this formula for the sum of the cubes of $n$ natural numbers:

$(1^{3}+2^{3}+3^{3}+.....+n^{3})=\left[\frac{n(n+1)}{2}\right]^2$

Solution:

Given:

$14^{3}+16^{3}+18^{3}+.....+30^{3}$

$= 2^{3}\times7^{3}+2^{3}\times8^{3}+2^{3}\times9^{3}+.....+2^{3}\times15^{3}$

$= 2^{3}(7^{3}+8^{3}+9^{3}+.....+15^{3})$

$= 2^{3}[(1^{3}+2^{3}+3^{3}+4^{3}+....+15^{3})-(1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3})]$

The sum of cubes of $n$ natural numbers$ =\left[\frac{n(n+1)}{2}\right]^2$

For $n = 6$, we have

$1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}=\left[\frac{6(6+1)}{2}\right]^2 = \left[\frac{42}{2}\right]^2 = (21)^{2} =$ 441

For $n = 15$, we have

$1^{3}+2^{3}+3^{3}+4^{3}+....+15^{3}=\left[\frac{15(15+1)}{2}\right]^2 = \left[\frac{240}{2}\right]^2 = (120)^{2} =$ 14400

Now, $2^{3}[(1^{3}+2^{3}+3^{3}+4^{3}+....+15^{3})-(1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3})]$

$=2^{3}(14400 - 441)$

$= 8 × 13959$

$= 111672$

Hence, the correct answer is option (4).

Q.5.

If Un = $\frac{1}{n}-\frac{1}{n+1}$, then the value of U1 + U2 + U3 + U4 + U5 is:

1. $\frac{1}{4}$

2. $\frac{5}{6}$

3. $\frac{1}{6}$

4. $\frac{1}{3}$

Hint: You have to put values of n, that is 1, 2, 3, 4, 5 in the first equation to solve it.

Solution:

Un = $\frac{1}{n}-\frac{1}{n+1}$

So, U1 = $\frac{1}{1}-\frac{1}{1+1}=1-\frac{1}{2}$

Similarly, U2 = $\frac{1}{2}-\frac{1}{3}$

U3 = $\frac{1}{3}-\frac{1}{4}$

U4 = $\frac{1}{4}-\frac{1}{5}$

U5 = $\frac{1}{5}-\frac{1}{6}$

Adding all this we get,

U1 + U2 + U3 + U4 + U5 = $1-\frac{1}{6}=\frac{5}{6}$

Hence, the correct answer is option (2).

Q.6.

If $\small 1^{2}+2^{2}+3^{2}+......+\ p^{2}=\frac{p(p+1)(2p+1)}{6}$, then $\small 1^{2}+3^{2}+5^{2}+......+17^{2}$ is equal to:

1. 1785

2. 1700

3. 980

4. 969

Hint: Use the given formula and simplify.

$1^2+2^2+3^2+......+n^2=\frac{n(n+1)(2n+1)}{6}$

Solution:

Given: $1^{2}+2^{2}+3^{2}+......+p^{2}=\frac{p(p+1)(2p+1)}{6}$

So, $1^{2}+2^{2}+3^{2}+......+17^{2}=\frac{17(17+1)(2\times 17+1)}{6}$-----------------(1)

Also, $2^{2}+4^{2}+6^{2}+......+16^{2}=2^{2}(1^{2}+2^{2}+3^{2}+......+8^{2})$

$⇒2^{2}+4^{2}+6^{2}+......+16^{2}=2^{2}(\frac{8(8+1)(2\times 8+1)}{6})$----------------(2)

Subtracting equation (2) from equation (1), we get,

$1^{2}+3^{2}+5^{2}+......+17^{2}=\frac{17(17+1)(2\times 17+1)}{6}-2^{2}(\frac{8(8+1)(2\times 8+1)}{6})$

$\therefore1^{2}+3^{2}+5^{2}+......+17^{2}= 969$

Hence, the correct answer is option (4).

Q.7.

$\small \left ( 5^{2}+6^{2}+7^{2}+.....+10^{2} \right )$ is equal to:

1. 330

2. 345

3. 355

4. 360

Hint: Use formula: $1^{2}+2^{2}+3^{2}+.....+n^{2}=\frac{n(n+1)(2n+1)}{6}$ and solve.

Solution:

We know that $1^{2}+2^{2}+3^{2}+.....+n^{2}=\frac{n(n+1)(2n+1)}{6}$

So, $1^{2}+2^{2}+3^{2}+.....+10^{2}=\frac{10(10+1)(2\times 10+1)}{6}$-----------------(1)

Also, $1^{2}+2^{2}+.....+4^{2}=\frac{4(4+1)(2\times 4+1)}{6}$------------------------------(2)

Subtracting equation (2) from (1), we get,

$5^{2}+6^{2}+.....+10^{2}=\frac{10(10+1)(2\times 10+1)}{6}-\frac{4(4+1)(2\times 4+1)}{6}$

$\therefore5^{2}+6^{2}+.....+10^{2}=355$

Hence, the correct answer is option (3).

Q.8.

Which of the following statements is true?

I. $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\ldots \ldots \frac{1}{110}<\frac{5}{6}$

II. $\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\ldots \ldots \frac{1}{143}>\frac{7}{13}$

1. Only I

2. Both I and II

3. Only II

4. Neither I nor II

Hint: Solve the problem by expanding the LHS terms.

Solution:

Statement I:

$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\ldots \ldots \frac{1}{110}<\frac{5}{6}$

Expand LHS

$⇒\frac{1}{2}+\frac{1}{2\times{3}}+\frac{1}{3\times{4}}+\ldots \ldots \frac{1}{10\times{11}}<\frac{5}{6}$

$⇒\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots \ldots \frac{1}{10}-\frac{1}{11}<\frac{5}{6}$

$⇒1-\frac{1}{11}<\frac{5}{6}$

$⇒\frac{10}{11}<\frac{5}{6}$

which is wrong,

So, statement I is incorrect.

Statement II:

$\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\ldots \ldots \frac{1}{143}>\frac{7}{13}$

$⇒\frac{1}{3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\ldots \ldots \frac{1}{11\times13}>\frac{7}{13}$

$⇒\frac{1}{3}+\frac{2}{2}[\frac{1}{3\times5}+\frac{1}{5\times7}+\ldots \ldots \frac{1}{11\times13}]>\frac{7}{13}$

$⇒\frac{1}{3}+\frac{1}{2}[\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\ldots \ldots \frac{1}{11}-\frac{1}{13}]>\frac{7}{13}$

$⇒\frac{1}{3}+\frac{1}{2}[\frac{1}{3}-\frac{1}{13}]>\frac{7}{13}$

$⇒\frac{1}{3}+\frac{5}{39}>\frac{7}{13}$

$⇒\frac{6}{13}>\frac{7}{13}$

which is wrong

So, statement II is incorrect.

$\therefore$ Both the statement is incorrect.

Hence, the correct answer is option (4).

Q.9.

What is the sum of the first 20 terms of the following series?

$1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \ldots$

1. 3160

2. 2940

3. 3240

4. 3080

Hint: Find the sum of the first 20 terms of the following series using formulas,

The sum of the first $n$ consecutive number = $\frac{n(n+1)}{2}$

The sum of the square of the first $n$ consecutive number = $\frac{n(n + 1) (2n + 1)}{6}$.

Solution:

Given: The series is $1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \ldots$.

The sum of the first $n$ consecutive number = $\frac{n(n+1)}{2}$

The sum of the square of the first $n$ consecutive number = $\frac{n(n + 1) (2n + 1)}{6}$

The sum of the first 20 terms of the following series is given as,

$n$ the term = $n(n+1)$

Sum = $\sum n(n+1)=\sum (n^2+n)=\sum n^2+\sum n$

Sum = $\frac{n(n + 1) (2n + 1)}{6}+\frac{n(n+1)}{2}$

⇒ Sum = $\frac{n(n+1)}{2}[\frac{2n+1}{3}+1]$

⇒ Sum = $\frac{n(n+1)}{2}[\frac{2n+1+3}{3}]$

⇒ Sum = $\frac{n(n+1)(2n+4)}{6}$ (equation 1)

Substitute $n=20$ in the equation (1),

Sum = $\frac{20(20+1)(2\times20+4)}{6}$

⇒ Sum = $\frac{20\times 21\times44}{6}$ = $3080$

Hence, the correct answer is option (4).

Q.10.

What is the value of $\frac{7}{2}+\frac{11}{3}+\frac{7}{6}+\frac{11}{15}+\frac{7}{12}+\frac{11}{35}+\ldots \ldots+\frac{7}{156}+\frac{11}{575}$?

1. $\frac{3917}{355}$

2. $\frac{3816}{325}$

3. $\frac{3714}{345}$

4. $\frac{3216}{315}$

Hint: The total sum of the series = The sum of the first series + The sum of the second series

Determine the value of the series by solving two distinct series independently.

Solution:

Given: The series is $\frac{7}{2}+\frac{11}{3}+\frac{7}{6}+\frac{11}{15}+\frac{7}{12}+\frac{11}{35}+\ldots \ldots+\frac{7}{156}+\frac{11}{575}$.

Determine the value of the series by solving two distinct series independently.

The sum of the first series = $\frac{7}{2}+\frac{7}{6}+\frac{7}{12}+...+\frac{7}{156}$

⇒ Sum = $7[1–(\frac{1}{2})+(\frac{1}{2}–\frac{1}{3})+(\frac{1}{3}–\frac{1}{4})+...+(\frac{1}{12}–\frac{1}{13})]$

⇒ Sum = $7[1–\frac{1}{13}]$

⇒ Sum = $7\times\frac{12}{13}=\frac{84}{13}$

The sum of the second series = $\frac{11}{3}+\frac{11}{15}+\frac{11}{35}+...+\frac{11}{575}$

⇒ Sum = $11[\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{575}]$

⇒ Sum = $\frac{11}{2}[\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{575}]$

⇒ Sum = $\frac{11}{2}[1–(\frac{1}{3})+(\frac{1}{3}–\frac{1}{5})+(\frac{1}{5}–\frac{1}{7})+...+(\frac{1}{23}–\frac{1}{25})]$

⇒ Sum = $\frac{11}{2}[1–\frac{1}{25}]$

⇒ Sum = $\frac{11}{2}\times\frac{24}{25}=\frac{132}{25}$

The total sum of the series = The sum of the first series + The sum of the second series

The total sum of the series $=\frac{84}{13}+\frac{132}{25}$

$=\frac{84\times 25+132\times 13}{25\times 13}=\frac{2100+1716}{325}=\frac{3816}{325}$

Hence, the correct answer is option (2).