Progressions
Ankur Sinha, PhD
Production and Quantitative Methods
Indian Institute of Management
Ahmedabad
Preparatory Programme - Mathematics
Arithmetic Progression
1, 5, 9, 13, 17 …
First Term
Common Difference
a=1
d=4
Denote the terms in a general arithmetic sequence as:
t1, t2, t3, t4, t5 …
a = t1 and ti - ti-1 = d
Then what is the nth term in the sequence?
Preparatory Programme - Mathematics
nth Term
1, 5, 9, 13, 17 …
+4 +4 +4 +4
To reach the nth term
you need to add +4 to
the first term how many
times?
(n-1) times
Therefore, nth term in the following arithmetic sequence
t1, t2, t3, t4, t5 …
is given by: tn = a + (n-1) d
Preparatory Programme - Mathematics
Problems
P1: The first term of an A.P. series is 2 and the common difference
is 1. Find the 20th term.
P2: The 5-th and the 12-th term of an A.P. series are 14 and 35
respectively. Find the first term and the common difference.
P3: The sum of three numbers in A.P. is 15 and their product is 80;
find the numbers.
Preparatory Programme - Mathematics
Sum of First n Terms
The story is told of a school teacher in the 1700's who wanted to keep her class
busy while she graded papers so she asked them to add up all of the numbers
from 1 to 100. These numbers are an arithmetic sequence with common difference
1. Carl Friedrich Gauss was in the class and had the answer in a minute or two.
This is what he did:
sums to 101
sums to 101
1 + 2 + 3 + 4 + 5 + . . . + 96 + 97 + 98 + 99 + 100
sums to 101
sums to 101
With 100 numbers there are 50 pairs that add up to 101. Therefore, the
sum of the series is 50(101) = 5050
Preparatory Programme - Mathematics
Sum of First n Terms
For the following sequence,
t1, t2, t3, t4, t5 … tn
Let the first term be a = t1 and the last term be l = tn, then
the sum of the first n terms is given as
n
Sn = ( a + l )
2
or
n
Sn = ( 2a + (n −1)d )
2
Preparatory Programme - Mathematics
Problems
P4: Find the sum 72+70+68+…+40.
P5: How many terms of the series 93+90+87+… sum up
to 975? Also find the last term.
Preparatory Programme - Mathematics
Problems
P6: A company engaged in the production of computers produced
300 units in the third year and 700 units in the seventh year.
Assuming that production increases uniformly by a fixed number
every year, find:
1.  the production in the first year.
2.  the total production in seven years.
3.  the production in the tenth year.
Preparatory Programme - Mathematics
Problems
P7: On 1st January every year, a man buys NSC (National Savings
Certificate) of value exceeding that of his last year’s purchase by
Rs.100. After 10 years he finds that the total value of the certificates
held by him is Rs.54,500. Find the value of the certificates
purchased by him
1.  in the first year.
2.  in the eighth year.
Preparatory Programme - Mathematics
Geometric Progression
2, 6, 18, 54, 162 …
Dividing two consecutive terms, ti/ti-1 always gives us the same ratio
3. Therefore, this series is in a geometric progression with common
ratio 3.
First Term
Common Ratio
a=2
r=3
How do you get the nth term of this sequence?
Preparatory Programme - Mathematics
nth Term
2, 6, 18, 54, 162 …
×3 ×3 ×3 ×3
To reach the nth term
you need to multiply the
first term by 3 how many
times?
(n-1) times
Therefore, nth term in the following geometric sequence
t1, t2, t3, t4, t5 …
is given by: tn = arn-1
Preparatory Programme - Mathematics
Problems
P8: Find the eighth and tenth terms of the G.P. series 2, 4, 8, 16 …
P9: Which term of the following series is 576? 4.5, 9, 18, …
P10: Find three numbers in G.P. whose sum is 26 and the
product is 216.
P11: Find the value of k if k−3, 2k−3 and 4k+3 are in G.P.
Preparatory Programme - Mathematics
Sum of First n Terms
2 + 6 + 18 + 54 + 162 = Sn
− ( 2 + 6 + 18 + 54 + 162 = Sn ) × 3
2 − 0 − 0 − 0 − 0 − 486 = (1 − 3) × Sn
Therefore, Sn = 242
Computation of the sum of the series: 2 + 6 + 18 + 54 + 162
The sequence is multiplied term by term by 3, and then subtracted
from the original sequence. Doing this only two terms remain.
Preparatory Programme - Mathematics
Sum of First n Terms
General formula for the sum of first n terms in a
geometric progression
1− r n
Sn = a
, r ≠1
1− r
Preparatory Programme - Mathematics
Problems
P12: Find the sum of the following series: 6 + 12 + 24 + … + 1536
P13: A man borrows Rs.19,682 without interest and agrees to repay
in 9 monthly installments, each installment being thrice the
preceding one. After the seventh installment, he wants to repay the
balance in lump. How much he will have to pay now?
Preparatory Programme - Mathematics
Infinite Series
1 1 1 1
+ + + +
2 4 8 16
1
2
0
Find the sum of this geometric
series up to infinity
+
1
4
1 1
+
+
8 16
Addition of each term makes you converge towards 1.
However, the sum never reaches 1 for finite terms.
1
Preparatory Programme - Mathematics
Infinite Series
For an infinite series of the form ark-1 the sum up to
infinity is given as follows:
a
a + ar + ar + ar + ar +…+ ∞ =
1− r
2
3
4
For an infinite series of the form ark-1 the sum up to
infinity is given as follows:
a
a + 2ar + 3ar + 4ar +…+ ∞ =
(1− r)2
2
3
Derive it!
Note that: −1 < r < 1
Preparatory Programme - Mathematics
Problems
P14: Find the sum of the following series
2 3 2 3 2 3
+ 2 + 3 + 4 + 5 + 6 ++ ∞
5 5 5 5 5 5
P15: Find the sum of the following series
4 + 4 + 3+ 2 +
20 24
+ ++ ∞
16 32
P16: Find the sum of the following series
2
3
"1%
"1%
1
2 + 3× 2 × + 4 × 2 × $ ' + 5 × 2 × $ ' ++ ∞
#2&
#2&
2
Preparatory Programme - Mathematics
Special Progressions
n
n
Sn = ∑ i = 1+ 2 + 3+ 4 + 5 +…+ n = (n +1)
2
i=1
n
n
Sn = ∑ i = 1 + 2 + 3 + 4 + 5 +…+ n = (n +1)(2n +1)
6
i=1
2
2
2
2
2
2
2
n
"n
%
Sn = ∑ i = 1 + 2 + 3 + 4 + 5 +…+ n = $ (n +1)'
#2
&
i=1
3
3
3
3
3
3
2
3
Preparatory Programme - Mathematics
Additional Problems
P17: Find the sum of the first n terms of the following series:
i
Ti = (i +1)
2
1 + (1+2) + (1+2+3) + … + Tn
n
n
"i
%
Sn = ∑Ti = ∑$ (i +1)'
#
&
i=1
i=1 2
Hint:
n
i2 n i
Sn = ∑ +∑
i=1 2
i=1 2
P18: Find the sum to n terms of the series:
1 + (1+4) + (1+4+42) + (1+4+42+43) + … + Tn
1− r i
Ti = a
1− r
n
Hint:
n
# 1− r i &
Sn = ∑Ti = ∑% a
(
1−
r
$
'
i=1
i=1
n
n
a
ar i
Sn = ∑
−∑
1−
r
i=1
i=1 1− r
Preparatory Programme - Mathematics
Thank you
Preparatory Programme - Mathematics