# Sequence and Series

## Short Questions

Define sequence and series.

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.

Define arithmetic sequence.

Define geometric sequence.

Define arithmetic mean.

Define geometric mean.

Find the arithmetic mean and geometric mean between 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.

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.

4, a and 16 are in geometric sequence.

As, their common ratio(r) is equal,

a/4 = 16/a
or, a^{2} = 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.

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.

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 + m^{2} + 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,
t_{n} = 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,
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

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,
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} = $\frac{\text{a(1 - }r^n)}{\text{1 - r}}$
= $\frac{\text{9{1 - } \left(\frac{1}{3}\right)^n } }{\text{1 - }\frac{1}{3}}$

= $\frac{27}{2} \left{\text{1 - }\frac{1}{3^6}\right}$

= $\frac{27}{2} \left{ \frac{728}{729} \right}$

= $\frac{364}{27}$

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/r^{2}, a/r, a, ar, ar^{2}

Given, Third term = 3 = a ---- (A)

Now, Product of first five terms is,

= a/r^{2}× a/r× a× ar× ar^{2}

= $\frac{a^5 \text{ × } r^3}{r^3}$

= a^{5}

= 3^{5} [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.

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 = 3p^{2} ---- (A)

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

or, 9p^{4} = 9

or, p^{4} = 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.

First term (a) = 2

Third term (t

_{3}) = 242

Now,
t_{3} = 242
or, ar^{3-1} = 242

or, 2 × r^{2} = 242

or, r^{2} = 121

∴ r = ±11

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

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,

S_{8} = $\frac{a(r^n \text{ - 1})}{\text{r - 1}}$
= $\frac{2(2^8 \text{ - 1})}{\text{2 - 1}}$

= 2(256 - 1)

= 510