Question 1

Exercises 1 to 6 whose nth terms are:

1. an = n(n + 2)
2. an = n/(n + 1)
3. an = 2^n
4. an = (2n - 3)/6
5. an = (-1)^{n-1} * 5^{n+1}
6. an = n * (n^2 + 5)/4

Find the indicated terms in each of the sequences in Exercises 1 to 6.

Accepted Answer

For each sequence given by the nth term an, substitute the required values of n to find the indicated terms.

Example for sequence 1: an = n(n + 2)
- To find a1, substitute n=1: a1 = 1*(1+2) = 3
- To find a2, substitute n=2: a2 = 2*(2+2) = 8

Similarly, for each sequence, substitute the given n values to find the terms. — The nth term formula gives the value of the term at position n. By substituting the value of n, the term can be calculated directly.

For example, for sequence 3: an = 2^n
- a1 = 2^1 = 2
- a2 = 2^2 = 4
- a3 = 2^3 = 8

This method applies to all sequences given.

Question 2

Write the first five terms of each of the sequences whose nth terms are:
(i) a_n = 4n - 3
(ii) a_n = \frac{n^2}{2^n}
(iii) a_n = (-1)^{n-1} n^3
(iv) a_n = \frac{n(n - 2)}{n + 3}

Accepted Answer

Solution:
(i) a_n = 4n - 3
First five terms:
For n=1: a_1 = 4(1) - 3 = 4 - 3 = 1
For n=2: a_2 = 4(2) - 3 = 8 - 3 = 5
For n=3: a_3 = 4(3) - 3 = 12 - 3 = 9
For n=4: a_4 = 4(4) - 3 = 16 - 3 = 13
For n=5: a_5 = 4(5) - 3 = 20 - 3 = 17
So, first five terms are: 1, 5, 9, 13, 17

(ii) a_n = \frac{n^2}{2^n}
First five terms:
For n=1: a_1 = 1^2 / 2^1 = 1/2 = 0.5
For n=2: a_2 = 4 / 4 = 1
For n=3: a_3 = 9 / 8 = 1.125
For n=4: a_4 = 16 / 16 = 1
For n=5: a_5 = 25 / 32 = 0.78125
So, first five terms are: 0.5, 1, 1.125, 1, 0.78125

(iii) a_n = (-1)^{n-1} n^3
First five terms:
For n=1: a_1 = (-1)^{0} * 1^3 = 1 * 1 = 1
For n=2: a_2 = (-1)^{1} * 8 = -8
For n=3: a_3 = (-1)^{2} * 27 = 27
For n=4: a_4 = (-1)^{3} * 64 = -64
For n=5: a_5 = (-1)^{4} * 125 = 125
So, first five terms are: 1, -8, 27, -64, 125

(iv) a_n = \frac{n(n - 2)}{n + 3}
First five terms:
For n=1: a_1 = (1*(1-2)) / (1+3) = (1*(-1))/4 = -1/4 = -0.25
For n=2: a_2 = (2*(2-2)) / (2+3) = (2*0)/5 = 0
For n=3: a_3 = (3*(3-2)) / (3+3) = (3*1)/6 = 3/6 = 0.5
For n=4: a_4 = (4*(4-2)) / (4+3) = (4*2)/7 = 8/7 ≈ 1.1429
For n=5: a_5 = (5*(5-2)) / (5+3) = (5*3)/8 = 15/8 = 1.875
So, first five terms are: -0.25, 0, 0.5, 1.1429, 1.875 — To find the first five terms of a sequence given its nth term formula, substitute n = 1, 2, 3, 4, 5 into the formula and simplify each expression to get the terms.

Question 3

$a_1 = 3, a_n = 3a_{n-1} + 2$  for all  $n > 1$

Accepted Answer

Given the recurrence relation $a_1 = 3$, and $a_n = 3a_{n-1} + 2$ for $n > 1$, find the corresponding series.

Solution:
Start with $a_1 = 3$.
Calculate next terms:
$a_2 = 3a_1 + 2 = 3 	imes 3 + 2 = 11$
$a_3 = 3a_2 + 2 = 3 	imes 11 + 2 = 35$
$a_4 = 3a_3 + 2 = 3 	imes 35 + 2 = 107$
$a_5 = 3a_4 + 2 = 3 	imes 107 + 2 = 323$

Thus, the series is: 3, 11, 35, 107, 323, ... — Using the given recurrence relation, each term is computed from the previous term by multiplying by 3 and adding 2. Starting from $a_1=3$, subsequent terms are calculated stepwise.

Question 4

$a_1 = -1, a_n = \frac{a_{n-1}}{n}, n \geq 2$

Accepted Answer

Given $a_1 = -1$ and $a_n = \frac{a_{n-1}}{n}$ for $n \geq 2$, find the corresponding series.

Solution:
Start with $a_1 = -1$.
Calculate next terms:
$a_2 = \frac{a_1}{2} = \frac{-1}{2} = -\frac{1}{2}$
$a_3 = \frac{a_2}{3} = \frac{-\frac{1}{2}}{3} = -\frac{1}{6}$
$a_4 = \frac{a_3}{4} = \frac{-\frac{1}{6}}{4} = -\frac{1}{24}$
$a_5 = \frac{a_4}{5} = \frac{-\frac{1}{24}}{5} = -\frac{1}{120}$

Thus, the series is: $-1, -\frac{1}{2}, -\frac{1}{6}, -\frac{1}{24}, -\frac{1}{120}, ...$ — Each term is obtained by dividing the previous term by the term number $n$. Starting from $-1$, the terms become $-1, -\frac{1}{2}, -\frac{1}{6}, ...$ which are negative reciprocals of factorials.

Question 5

$a_1 = a_2 = 2, a_n = a_{n-1} - 1, n > 2$

Accepted Answer

Given $a_1 = a_2 = 2$ and $a_n = a_{n-1} - 1$ for $n > 2$, find the corresponding series.

Solution:
Start with $a_1 = 2$, $a_2 = 2$.
Calculate next terms:
$a_3 = a_2 - 1 = 2 - 1 = 1$
$a_4 = a_3 - 1 = 1 - 1 = 0$
$a_5 = a_4 - 1 = 0 - 1 = -1$
$a_6 = a_5 - 1 = -1 - 1 = -2$

Thus, the series is: 2, 2, 1, 0, -1, -2, ... — Starting from two initial terms equal to 2, each subsequent term is obtained by subtracting 1 from the previous term.

Question 6

The Fibonacci sequence is defined by

$$
1 = a _ {1} = a _ {2} 	ext{ and } a _ {n} = a _ {n - 1} + a _ {n - 2}, n > 2.
$$

Find  $\frac{a_{n + 1}}{a_n}$  for  $n = 1,2,3,4,5$

Accepted Answer

Given the Fibonacci sequence defined by $a_1 = a_2 = 1$ and $a_n = a_{n-1} + a_{n-2}$ for $n > 2$, find $\frac{a_{n+1}}{a_n}$ for $n=1,2,3,4,5$.

Solution:
First, list the first several terms:
$a_1 = 1$
$a_2 = 1$
$a_3 = a_2 + a_1 = 1 + 1 = 2$
$a_4 = a_3 + a_2 = 2 + 1 = 3$
$a_5 = a_4 + a_3 = 3 + 2 = 5$
$a_6 = a_5 + a_4 = 5 + 3 = 8$

Now compute the ratios:
For $n=1$: $\frac{a_2}{a_1} = \frac{1}{1} = 1$
For $n=2$: $\frac{a_3}{a_2} = \frac{2}{1} = 2$
For $n=3$: $\frac{a_4}{a_3} = \frac{3}{2} = 1.5$
For $n=4$: $\frac{a_5}{a_4} = \frac{5}{3} \approx 1.6667$
For $n=5$: $\frac{a_6}{a_5} = \frac{8}{5} = 1.6$

Thus, the values are: 1, 2, 1.5, 1.6667, 1.6. — Calculate the first six Fibonacci numbers, then find the ratio of consecutive terms for the specified values of $n$.

Question 7

Find the $20^{	ext{th}}$ and $n^{	ext{th}}$ terms of the G.P. $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots$

Accepted Answer

Given G.P. is $ \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots $

First term, $ a = \frac{5}{2} $
Common ratio, $ r = \frac{\frac{5}{4}}{\frac{5}{2}} = \frac{5}{4} \times \frac{2}{5} = \frac{1}{2} $

The $ n^{th} $ term of a G.P. is given by:
$$
T_n = a r^{n-1}
$$

Therefore,
$$
T_{20} = \frac{5}{2} \times \left( \frac{1}{2} \right)^{19} = \frac{5}{2} \times \frac{1}{2^{19}} = \frac{5}{2^{20}}
$$

and

$$
T_n = \frac{5}{2} \times \left( \frac{1}{2} \right)^{n-1} = \frac{5}{2^n}
$$ — Step 1: Identify first term and common ratio.
Step 2: Use formula for nth term of G.P.: $ T_n = a r^{n-1} $.
Step 3: Substitute values to find $ T_{20} $ and general $ T_n $.
Step 4: Simplify the expressions.

Question 8

Find the $12^{	ext{th}}$ term of a G.P. whose $8^{	ext{th}}$ term is 192 and the common ratio is 2.

Accepted Answer

Given:
$$ T_8 = 192, \quad r = 2 $$

We know the nth term of a G.P. is:
$$
T_n = a r^{n-1}
$$

So,
$$
T_8 = a r^{7} = 192
$$
$$
a \times 2^{7} = 192
$$
$$
a \times 128 = 192 \implies a = \frac{192}{128} = \frac{3}{2}
$$

Now, find $ T_{12} $:
$$
T_{12} = a r^{11} = \frac{3}{2} \times 2^{11} = \frac{3}{2} \times 2048 = 3 \times 1024 = 3072
$$ — Step 1: Use given $ T_8 $ and $ r $ to find first term $ a $.
Step 2: Use $ a $ and $ r $ to find $ T_{12} $.
Step 3: Calculate powers of 2 and multiply accordingly.

Question 9

The $5^{	ext{th}}$, $8^{	ext{th}}$ and $11^{	ext{th}}$ terms of a G.P. are $p$, $q$ and $s$, respectively. Show that $q^2 = ps$.

Accepted Answer

Let the first term be $ a $ and common ratio be $ r $.

Then,
$$
T_5 = a r^{4} = p
$$
$$
T_8 = a r^{7} = q
$$
$$
T_{11} = a r^{10} = s
$$

Calculate $ q^2 $:
$$
q^2 = (a r^{7})^2 = a^2 r^{14}
$$

Calculate $ p s $:
$$
p s = (a r^{4})(a r^{10}) = a^2 r^{14}
$$

Since $ q^2 = p s $, the relation is proved. — Step 1: Express terms in terms of $ a $ and $ r $.
Step 2: Compute $ q^2 $ and $ p s $.
Step 3: Show both are equal, proving the relation.

Question 10

The $4^{	ext{th}}$ term of a G.P. is square of its second term, and the first term is $-3$. Determine its $7^{	ext{th}}$ term.

Accepted Answer

Given:
$$ a = -3 $$
Let common ratio be $ r $.

Given:
$$
T_4 = (T_2)^2
$$

Express terms:
$$
T_4 = a r^{3} = -3 r^{3}
$$
$$
T_2 = a r = -3 r
$$

So,
$$
-3 r^{3} = (-3 r)^2 = 9 r^{2}
$$

Divide both sides by $ r^{2} $ (assuming $ r \neq 0 $):
$$
-3 r = 9 \implies r = -3
$$

Now, find $ T_7 $:
$$
T_7 = a r^{6} = -3 \times (-3)^6 = -3 \times 729 = -2187
$$ — Step 1: Use given relation between $ T_4 $ and $ T_2 $ to find $ r $.
Step 2: Substitute $ r $ and $ a $ to find $ T_7 $.
Step 3: Calculate powers and multiply.

Question 11

Which term of the following sequences:

(a) $2,2\sqrt{2},4,\ldots$ is 128?

(b) $\sqrt{3}, 3, 3\sqrt{3}, \ldots$ is 729?

(c) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$?

Accepted Answer

(a) Sequence: 2, 2√2, 4, ...

First term $ a = 2 $, common ratio:
$$
r = \frac{2\sqrt{2}}{2} = \sqrt{2}
$$

Find $ n $ such that $ T_n = 128 $:
$$
T_n = a r^{n-1} = 2 (\sqrt{2})^{n-1} = 128
$$
Divide both sides by 2:
$$
(\sqrt{2})^{n-1} = 64
$$
Recall $ \sqrt{2} = 2^{1/2} $, so:
$$
(2^{1/2})^{n-1} = 2^{(n-1)/2} = 2^{6} \implies \frac{n-1}{2} = 6 \implies n-1 = 12 \implies n = 13
$$

(b) Sequence: $ \sqrt{3}, 3, 3\sqrt{3}, \ldots $

First term $ a = \sqrt{3} = 3^{1/2} $, common ratio:
$$
r = \frac{3}{\sqrt{3}} = \sqrt{3} = 3^{1/2}
$$

Find $ n $ such that $ T_n = 729 $:
$$
T_n = a r^{n-1} = 3^{1/2} (3^{1/2})^{n-1} = 3^{1/2 + (n-1)/2} = 3^{(n)/2} = 729
$$
Since $ 729 = 3^{6} $,
$$
3^{n/2} = 3^{6} \implies \frac{n}{2} = 6 \implies n = 12
$$

(c) Sequence: $ \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots $

First term $ a = \frac{1}{3} = 3^{-1} $, common ratio:
$$
r = \frac{1/9}{1/3} = \frac{1}{9} \times 3 = \frac{1}{3} = 3^{-1}
$$

Find $ n $ such that $ T_n = \frac{1}{19683} $:
$$
T_n = a r^{n-1} = 3^{-1} (3^{-1})^{n-1} = 3^{-1 - (n-1)} = 3^{-n} = \frac{1}{19683}
$$
Since $ 19683 = 3^{9} $,
$$
3^{-n} = 3^{-9} \implies n = 9
$$ — For each sequence:
- Identify first term $ a $ and common ratio $ r $.
- Use formula $ T_n = a r^{n-1} $.
- Express given term as power of base.
- Equate exponents and solve for $ n $.

Question 12

For what values of $x$, the numbers $-\frac{2}{7}, x, -\frac{7}{2}$ are in G.P.?

Accepted Answer

Given three numbers in G.P.:
$$
-\frac{2}{7}, x, -\frac{7}{2}
$$

For three numbers $ a, b, c $ to be in G.P.,
$$
b^2 = a c
$$

So,
$$
x^2 = \left(-\frac{2}{7}\right) \times \left(-\frac{7}{2}\right) = \frac{2}{7} \times \frac{7}{2} = 1
$$

Therefore,
$$
x^2 = 1 \implies x = \pm 1
$$ — Step 1: Use property of G.P. that middle term squared equals product of first and third.
Step 2: Substitute given values.
Step 3: Solve for $ x $.

Question 13

0.15, 0.015, 0.0015, ... 20 terms.

Accepted Answer

This is a geometric progression (G.P.) with first term a = 0.15 and common ratio r = 0.1 (since 0.015 / 0.15 = 0.1). The question likely asks to find the sum of 20 terms.

Sum of n terms of G.P. is S_n = a(1 - r^n) / (1 - r).

Here, n = 20, a = 0.15, r = 0.1.

Calculate r^20 = (0.1)^20 = 1e-20 (very small, approx 0).

So,
S_20 = 0.15 * (1 - 1e-20) / (1 - 0.1) = 0.15 / 0.9 = 1/6 = 0.166666... approximately.

Hence, sum of 20 terms is approximately 0.1667. — Identify the sequence as a G.P. with first term 0.15 and common ratio 0.1. Use the formula for sum of n terms of G.P. Substitute values and calculate the sum.

Question 14

$\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \ldots, n$ terms.

Accepted Answer

Given sequence: $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \ldots$

Check if it is a G.P.:

Second term / first term = $\frac{\sqrt{21}}{\sqrt{7}} = \sqrt{3}$

Third term / second term = $\frac{3\sqrt{7}}{\sqrt{21}} = \frac{3\sqrt{7}}{\sqrt{7 \times 3}} = \frac{3\sqrt{7}}{\sqrt{7} \sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$

So common ratio $r = \sqrt{3}$.

First term $a = \sqrt{7}$.

Sum of n terms:

$$ S_n = a \frac{r^n - 1}{r - 1} = \sqrt{7} \frac{(\sqrt{3})^n - 1}{\sqrt{3} - 1} $$

This formula can be used to find the sum or any term as needed. — Identify the sequence as a G.P. with first term $\sqrt{7}$ and common ratio $\sqrt{3}$. Use the sum formula for G.P. to find sums or terms.

Question 15

$1, -a, a^2, -a^3, \ldots, n$ terms (if $a \neq -1$).

Accepted Answer

Given sequence: 1, -a, a^2, -a^3, ...

This is a G.P. with first term $a_1 = 1$ and common ratio $r = -a$.

Sum of n terms:

$$ S_n = \frac{1 - (-a)^n}{1 - (-a)} = \frac{1 - (-1)^n a^n}{1 + a} $$

This formula is valid if $a \neq -1$ to avoid division by zero. — Identify the sequence as a G.P. with first term 1 and common ratio -a. Use the sum formula for G.P. with substitution.

Sequences and Series

Sequences and Series — Study Notes

8.1 Introduction

8.2 Sequences

8.3 Series

Practice Questions — Sequences and Series

All 14 Chapters in Mathematics