SSC math homework- Arithmatic progression and geometric progression

CHAPTER – 1

ARITHMETIC PROGRESSION AND

GEOMETRIC PROGRESSION

Sequence:

A sequence is a collection (lists) of numbers arranged in a definite order

according to some definite rule.

Each number in the sequence is called a term of the sequence.

(Textual Solved Examples)

Ex.1 Find the first four terms of the sequence whose ‘nth’ term is 3n + 1

Solution:

Here, t2 = 3n +1

For n = 1, t4 = 3 × 1 + 1 = 4

For n = 2, t8 = 3 × 2 + 1 = 7

For n = 3, t: = 3 × 3 + 1 = 10

For n = 4, t< = 3 × 4 + 1 = 13

Thus the first four terms of the sequence are 4, 7, 10, 13.

EXERCISE – 1.1

Textual

Solutions

1. For each sequence, find the next four terms :

i) 1, 2, 4, 7, 11, …

Find : t6 , t7 , t8 , t9.

Solution :

t1 = 1 , t2 = 2 , t3= 4 , t4 = 7 , t5 = 11

t2 − t1 = 2−1 = 1

t3 − t2 = 4 − 2 = 2

t4−t3 = 7 − 4 = 3

t5 − t4 = 11 − 7 = 4

∴ t6 = 11 + 5 = 16

t7 = 16 + 6 = 22

t8 = 22 + 7 = 29

t9 = 29 + 8 = 37

∴ The next four terms of the sequence are are 16 , 22 , 29 , 37.

*****

EXERCISE – 1.2

Textual

Solutions

1. Which of the following lists of numbers are Arithmetic Progressions?

Justify.

i) 1, 3, 6, 10, …

Find : If the numbers are Arithmetic Progression (A.P)

Solution :

t4 = 1, t8 = 3, t: = 6, t< = 10

d = t8 − t4 = 3 − 1 = 2

d = t: − t8 = 6 − 3 = 3

The difference (d) is not constant

∴ The given sequence is not an A. P.

EXERCISE – 1.3

Solutions

1. Find the twenty fifth term of the A.P : 12, 16, 20, 24, …

Solution :

Here a = 12

Hence, t4 = 12

t8 − t4 = 16 − 12 = 4

t: − t8 = 20 − 16 = 4

t< − t: = 24 − 20 = 4

∴ d = 4

t2 = a + (n − 1) d

t8J = 12 + (25 − 1)4

t8J = 12 + 24 × 4

t8J = 12 + 96

t8J = 108

∴ The 25th term of the A.P. is 108 .

EXERCISE – 1.4

Textual

Solutions

1. Find the sum of the first ‘n’ natural numbers and hence find the sum of

first 20 natural numbers.

Solution :

Here, t4 = 1, t2 = 20

S2 =

2

8

× (t4 + t2)

S2 =

8K

8

× (1 + 20)

S2 = 10 × 21

∴ LM = 210

EXERCISE – 1.5

Textual

Solutions

1. Find four consecutive terms in an A.P. whose sum is 12 and the sum of

3rd and 4th term is 14.

Solution :

Given : sum of 4 consecutive terms = 12

Sum of 3rd and 4th terms = 14

Find : Four consecutive terms in an A.P.

Let the four consecutive terms be a – 3d, a − d, a + d, a + 3d.

According to the 1st condition.

∴ (a − 3d) + (a − d) + (a + d) + (a + 3d) = 12

a + a + a + a - 3a – d + d + 3d =12

∴ 4a = 12

∴ a =

48

< 

∴ a = 3

According to the 2nd condition.

∴ (a + d) + (a + 3d) = 14

∴ 2a +4d = 14

∴ a +2d = 7 --- (dividing all by 2)

∴ 3 +2d = 7 --- (∵ a = 3)

∴ 2d = 7 − 3

∴ 2d = 4

∴ d =

< 

8

= 2

∴ d = 2

By putting, a = 3 and d = 2 we get four consecutive terms

a − 3d = 3 − 3× 2 = 3 − 6 = −3

a − d = 3 − 2 = 1

a + d = 3 + 2 = 5

a + 3d = 3 + 3× 2 = 3 + 6 = 9

∴ The four consecutive terms are −3 , 1, 5, 9.

*****

EXERCISE – 1.6

Textual

Solutions

1. Mary got a job with a starting salary of Rs. 15000/- per month. She will

get an incentive of Rs. 100/- per month. What will be her salary after 20

months.

Solution :

Given : Starting salary of Mary= Rs. 15000 per month.

Incentive = Rs. 100 per month.

Find : Mary’s salary after 20 months. (TUV)

Incentive per month = 100

salary is an A.P. = t4 = a = 15000

d = 100

t2 = a + (n − 1) d

t8K = 15000 + (20 − 1) 100

= 15000 + 1900

∴ t8K = 16900

∴ Mary’s salary after 20 months will be Rs. = 16,900 .

EXERCISE – 1.7

Textual

Solutions

1. Find the ninth term of the G.P. 3, 6, 12, 24, …

Find : The ninth term of the G.P.

Solution :

Hence we have a = 3, n = 9, r = 2

Now t2 = ar2 W4

tX = 3 × 2X W4

tX = 3 × 2Y

tX = 3 × 256

TZ = 768

∴ The ninth term of the G.P. is 768.

*****

EXERCISE – 1.8

Textual

Solutions

1. Find the indicated sums for the following Geometric Progressions.

i) 2, 6, 18, … Find L[.

Solution :

Given : In a G.P. 2, 6, 18, …

Find : L[.

Here we have a = 1, r = 3 and 3 > 1

S2 =

](^_ W 4)

^W4

∴ S` =

8( :a W 4)

:W4

∴ S` =

8(84Y`W 4)

8

=

8 × 84Yb

8

=

<:`8

8

= 2186

∴ L[ = 2186

EXERCISE – 1.9

Textual

Solutions

1. Find three consecutive terms in a G.P. such that the sum of the first two

terms is 9 and the product of all the three is 216.

Find : Find three consecutive terms in a G.P.

Solution :

Let the three terms be

]

^

, a, ar.

Then

]

^

+ a + ar = 9 --- (1)

and

]

^

× a × ar = 216 --- (2)

From equation(2)

a: = 216

a = 6 ---(taking cube roots on both sides)

Substituting this value of a = 6 in equation (1) we have

b

^

+ 6 = 9

∴ 6 + 6r = 9r

∴ 9r − 6r = 6

∴ 3r = 6

∴ r =

b

:

∴ r = 2

∴ 1st term =

]

^

=

b

8

= 3

∴ 2nd term = a = 6

∴ 3rd term = ar = 6 × 2 = 12

∴ Thus the three terms in a G.P. are 3, 6, 12.

*****

PROBLEM SET – 1

Textual

Solutions

1. Find Tdd from the following A.P. 4, 9, 14, …

Find : Tdd

Solution :

t4 = 4, t8 = 9, t: = 14

d = t8 − t4 = 9 − 4 = 5

t: − t8 = 14 − 9 = 5

∴ d = 5, a = 4

t2 = a + (n − 1) d

t44 = 4 + (11 − 1) 5

t44 = 4 + 10 × 5

t44 = 4 + 50

∴ Tdd = 54