# Sequence and Series

## Short Questions

Define 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 Sn.
1 + 3 + 5 + ... + n, 2 + 4 + 8 + ... + n are the example of series.

Define arithmetic sequence.

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

Define geometric 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.

Define arithmetic mean.

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

Define geometric mean.

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

Find the arithmetic mean and geometric mean between 4 and 16.

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

Find the arithmetic mean and geometric mean between 3 and 27.

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

If 4, a and 16 are in the geometric sequence, find the value of a.

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

If the arithmetic mean between 4 and x is 34, find the their geometric mean.

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

If the geometric mean of 2 and x is 4, find arithmetic mean.

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

If m + 2, m + 8 and 17 + m are in geometric sequence, find the value of m.

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 ]

Which term of the series 5 + 9 + 13 + ... is 85?

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

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

Which term of the series 3 + 6 + 12 + ... is 192?

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

Now, tn = 192 or, arn-1 = 192
or, 3 × rn - 1 = 192
or, rn - 1 = 64
pr, (2)n - 1 = (2)6
or, n - 1 = 6
∴ n = 7

So, 192 is in seventh term.

Find the sum of the series 9 + 3 + 1 + ... + 1/27.

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

Now, tn = 1/27 or, arn - 1 = 1/27
or, 9 × rn - 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,
Sn = $\frac{\text{a(1 - }r^n)}{\text{1 - r}}$ = $\frac{\text{9{1 - } \left(\frac{1}{3}\right)^n } }{\text{1 - }\frac{1}{3}}$
= $\frac{27}{2} \left{\text{1 - }\frac{1}{3^6}\right}$
= $\frac{27}{2} \left{ \frac{728}{729} \right}$
= $\frac{364}{27}$

If the third terms of a geometric series is 3, find the product of first five terms.

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

Find the arithmetic mean and geometric mean of 3 and 27.

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

If 1/3, p, q, 9 are in the geometric sequence, find the values of p and q.

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

Find the common ratio of a geometric sequence, whose first term is 2 and the third term is 242.

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

Now, t3 = 242 or, ar3-1 = 242
or, 2 × r2 = 242
or, r2 = 121
∴ r = ±11

If 5, x, y, 11 is an arithmetic series. Find the value of x and y.

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

Find the sum of the this series: 2 + 4 + 8 + ... upto 8 terms.

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