Sequence and Series

Sequence: A set of numbers which is formed under a definite mathematical rule is called a sequence.
1, 3, 5, ..., ..., and, 2, 4, 8, ..., ... are the example of sequence.

Series: The sum of the terms of a sequence is called a series. The sum of the terms of the series is denoted by S_n.
1 + 3 + 5 + ... + n, 2 + 4 + 8 + ... + n are the example of series.

A sequence having same difference between the successive terms is called arithmetic sequence.

The sequence in which the ratio of any term to the preceding term is constant is called geometric sequence. The constant is called the common ratio.

In an arithmetic sequence, the term between the first term and last term is called arithmetic mean.

In geometric sequence the term between the first and last term is called geometric mean.

Given numbers are 4 and 16.
Arithmetic mean (A.M) = ?
Geometric mean (G.M) = ?

Now, We have,
A.M = $\frac{4 \text{ + } 16}{2}$ = 10
G.M = $\sqrt{4 \text{ × } 16}$ = 8

Given numbers are 3 and 27.
Arithmetic mean (A.M) = ?
Geometric mean (G.M) = ?

Now, We have,
A.M = $\frac{3 \text{ + } 27}{2}$ = 15
G.M = $\sqrt{3 \text{ × } 27}$ = 9

Here,
4, a and 16 are in geometric sequence.

As, their common ratio(r) is equal,
a/4 = 16/a or, a² = 64
∴ a = 8

Another method

Here,
4, a and 16 are in geometric sequence.
First term (a) = 4
Last term (b) = 16
Geometric mean (GM) = a

Now, We know,
GM = $\sqrt{ab}$ or, a = $\sqrt{4 × 16}$
or, a = $\sqrt{64}$
∴ a = 8

Here,
Given numbers are 4 and x
And, Arithmetic mean (A.M) = 34
Geometric mean (G.M) = ?

Now, We know, A.M = 34 or, (4 + x)/2 = 34
or, 4 + x = 68
So, x = 64

Finally, G.M = $\sqrt{4 × x}$ = $\sqrt{4 × 64}$
= 16

Solution,
Given numbers are 2 and x.
And, Geometric mean (G.M) = 4
Arithmetic mean (A.M) = ?

Now, We know, G.M = 4
or, $\sqrt{2x}$ = 4
Squaring on both side, we get,
or, 2x = 16
∴ x = 8

Finally, A.M = (2 + x)/2 = 10/2
= 5

Here,
m + 2, m + 8 and 17 + m are in geometric sequence.

As their common ratio are equal, $\frac{\text{m + 8}}{\text{m + 2}}$ = $\frac{\text{17 + m}}{\text{m + 8}}$ or, (m + 8)² = 17m + m² + 34 + 2m
or, 19m - 16m = 64 - 34
or, 3m = 30
∴ m = 10

Another method

You can also solve this by using the formula G.M = $\sqrt{ab}$ [as solved in the another method of this question ]

Here,
Given series is, 5 + 9 + 13 + ... ...
Where, first term (a) = 5
Common difference (d) = 9 - 5 = 4

Now, t_n = 85 or, a + (n - 1)d = 85
or, 5 + (n - 1)4 = 85
or, n - 1 = 80/4
∴ n = 21

Here,
Given series is, 3 + 6 + 12 + ...
Where, First term (a) = 3
Common ratio (r) = 6/3 = 2

Now, t_n = 192 or, ar^n-1 = 192
or, 3 × r^{n - 1} = 192
or, r^{n - 1} = 64
pr, (2)^{n - 1} = (2)⁶
or, n - 1 = 6
∴ n = 7

So, 192 is in seventh term.

Solution,
Given series, 9 + 3 + 1 + ... + 1/27
Where, first term (a) = 9
Common ratio (r) = 3/9 = 1/3

Now, t_n = 1/27 or, ar^{n - 1} = 1/27
or, 9 × r^{n - 1} = 1/27
or, $\left(\frac{1}{3}\right)^{\text{n - 1}}$ = $\left(\frac{1}{3}\right)^{5}$
∴ n = 6

Finally, Sum of given series,
S_n = $\frac{\text{a(1 - }r^n)}{\text{1 - r}}$ = $\frac{\text{9{1 - } \left(\frac{1}{3}\right)^n } }{\text{1 - }\frac{1}{3}}$
= $\frac{27}{2} \left{\text{1 - }\frac{1}{3^6}\right}$
= $\frac{27}{2} \left{ \frac{728}{729} \right}$
= $\frac{364}{27}$

Here,
Let, first five terms of geometric series are a/r², a/r, a, ar, ar²
Given, Third term = 3 = a ---- (A)

Now, Product of first five terms is,
= a/r²× a/r× a× ar× ar²
= $\frac{a^5 \text{ × } r^3}{r^3}$
= a⁵
= 3⁵ [From (A)]
= 243

Solution,
Given numbers are 3 and 27
Arithmetic mean (A.M) = ?
Geometric mean (G.M) = ?

Now, We have, A.M = $\frac{3 \text{ + } 27}{2}$ = 15
G.M = $\sqrt{3 \text{ × } 27}$ = 9

Solution,
1/3, p, q, 9 are in the geometric sequence

As its common ratio is equal, $\frac{p}{\frac{1}{3}}$ = $\frac{q}{p}$ = $\frac{9}{q}$ or, 3p = $\frac{q}{p}$ = $\frac{9}{q}$

Now, taking, 3p = $\frac{q}{p}$ or, q = 3p² ---- (A)

Again, taking, $\frac{q}{p}$ = $\frac{9}{q}$ or, $\frac{3p^2}{p}$ = $\frac{9}{3p^2}$
or, 9p⁴ = 9
or, p⁴ = 1
∴ p = 1

From (A),
∴ q = 3

Hence, p = 1 and q = 3

Here,
First term (a) = 2
Third term (t₃) = 242

Now, t₃ = 242 or, ar^3-1 = 242
or, 2 × r² = 242
or, r² = 121
∴ r = ±11

Here,
5, x, y, 11 is an arithmetic series

As common difference is equal, x - 5 = y - x = 11 - y

Taking, x - 5 = y - x or, 2x = 5 + y
or, x = (5 + y)/2 ---- (A)

Again, y - x = 11 - y or, 2y = 11 + x
or, 4y = 22 + 5 + y [From (A)]
or, 3y = 27
So, y = 9
And, x = 7 [From (A)]

Hence, x = 7 and y = 9

Solution,
Given series: 2 + 4 + 8 + ...
First term (a) = 2
Common ratio (r) = 4/2 = 2

Now, the Sum of the given series up to 8 terms,
S₈ = $\frac{a(r^n \text{ - 1})}{\text{r - 1}}$ = $\frac{2(2^8 \text{ - 1})}{\text{2 - 1}}$
= 2(256 - 1)
= 510