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
