ICSE Revision Notes for Progressions Class 10 Maths
Chapter Name 
Progressions 
Topics Covered 

Related Study 
The Concept of Arithmetic Progression
Let us look at the following example.
Example 1: Check whether the following sequences form an A.P. or not.
(i) 5√2  7√5, 6√2  6√5, 7√2  5√5, 8√2  4√5.....
(ii) 1/3, 5/6, 4/3, 11/6 .......
(iii) 1/(1 × 2), 1/(2 × 3), 1/(3 × 4), 1/(4 × 5) .....
(iv) A list of prime numbers greater than 2
(v) A list of the squares of natural numbers
Solution
(i) The given sequence is 5√2  7√5, 6√2  6√5, 7√2  5√5, 8√2  4√5........
Difference between the second and the first term = (6√2 – 6√5) – (5√2  7√5) = (√2 + √5)
Difference between the third and the second term = 7(√2  5√5) – (6√2  6√5) = √2 + √5x
Difference between the fourth and the third term = (8√2  4√5) – (7√2  5√5) = √2 + √5
Since the difference between any two consecutive terms is a constant, the sequence is an arithmetic progression.
(ii) The given sequence is 1/3, 5/6, 4/3, 11/6 ........
Difference between the second and the first term = 5/6 – 1/3 = 1/2
Difference between the third and the second term = 4/3 – 5/6 = 1/2
Difference between the fourth and the third term = 11/6 – 4/3 = 1/2
Since the difference between any two consecutive terms is a constant, the sequence is an arithmetic progression.
(iii) the given sequence is 1/(1 × 2), 1/(2 × 3), 1/(3 × 4), 1/(4 × 5) ......
This sequence can be rewritten as 1/2, 1/6, 1/12, 1/20.......
Difference between the second and the first term = 1/6  1/2 = 1/3
Difference between the third and the second term = 1/12 – 1/6 = 1/12
Difference between the fourth and the third term = 1/20 – 1/12 = 1/30
Since the difference between the consecutive terms is not a constant, the sequence is not an arithmetic progression.
(iv) The list of the prime numbers greater than 2 is: 3, 5, 7, 11 …
Difference between the second and the first term = 5 – 3 = 2
Difference between the third and the second term = 7 – 5 = 2
Difference between the fourth and the third term = 11 – 7 = 4
Since the difference between the consecutive terms is not a constant, the sequence is not an arithmetic progression.
(v) The list of the squares of natural numbers is: 1^{2}, 2^{2}, 3^{2}, 4^{2 }… = 1, 4, 9, 16 …
Difference between the second and the first term = 4 – 1 = 3
Difference between the third and the second term = 9 – 4 = 5
Difference between the fourth and the third term = 16 – 9 = 7
Since the difference between the consecutive terms is not a constant, the sequence is not an arithmetic progression.
The Terminologies Related to Arithmetic Progressions
An arithmetic progression (A.P.) is a sequence in which the difference between any two consecutive terms is constant.
Let us solve some more examples to understand this concept better.
Example 1: Find the common difference of the A.P. 2+√2, 2+2√2, 2+3√2, 2+4√2.... Also, state whether it is finite or infinite and write down its first term.
Answer
The given A.P. is 2 + √2, 2 + 2√2, 2 + 3√2, 2 + 4√2....
a_{1} = 2 + √2
a_{2} = 2 + 2√2
a_{3} = 2 + 3√2
a_{4} = 2 + 4√2
a_{2 }– a_{1} = (2 + 2√2) – (2 + √2) = √2
a_{3} – a_{2} = (2 + 3√2) – (2 + 2√2) = √2
∴ Common difference of the A.P. = d = √2
First term of the A.P. = a_{1 }= 2 + √2
The given A.P. is infinite because its last term cannot be calculated.
Example 2: State whether the following statements are true or false.
(i) The sequence √5  √6, 2√6,  √5 – 3√6, 2√5  4√6 ….. forms an A.P. with the first term as  √5  √6 and the common difference as .
(ii) The common difference of the A.P.1/3, 5/6, 4/3, 11/6 ….. is 1/2.
(iii) The n^{th }term of an A.P. is given by 3n – 4. Its common difference is 4.
Answer
(i) The given sequence is √5  √6, 2√6, √5  3√6, 2√5  4√6……
Difference between the first and the second term =  2√6 – (√5  √6) = √5  √6
Difference between the second and the third term =  √5  3√6 – (2√6) =  √5  √6
Difference between the third and the fourth term = 2√5  4v6 – ( √5  3√6) =  √5  √6
Since the difference between the consecutive terms of the sequence is constant, the given sequence is an A.P. with the first term as √5  √6 and the common difference as √5  √6.
Thus, the statement is false.
(ii) The given A.P. is 1/3, 5/6, 4/3, 11/6
Common difference of the A.P. = 5/6 – 1/3 = 4/3 – 5/6 = 1/6 – 4/3 = 1/2
Thus, the statement is true.
(iii) It is given that the nth term of the A.P. is given by 3n – 4.
Thus, first term of the A.P. = a = 3 × 1 – 4 = 3 – 4 = –1
Second term of the A.P. = a_{2} = 3 × 2 – 4 = 6 – 4 = 2
Third term of the A.P. = a_{3} = 3 × 3 – 4 = 9 – 4 = 5
Fourth term of the A.P. = a_{4} = 3 × 4 – 4 = 12 – 4 = 8
Thus, common difference of the A.P. = 8 – 5 = 5 – 2 = 2 – (–1) = 3 Thus, the statement is false.
nth Term of An Arithmetic Progression
We know what an arithmetic progression (A.P.) is. Also, we have learnt that there is a common difference between any two consecutive terms of an A.P..
Now, can we find the required term of a given A.P. with this information? Let us consider the A.P. 3, 7, 11, 15, ...
Here, first term (a) = 3 and common difference (d) = 4
Now, if we want to find the 5^{th }term of this A.P., then we will simply add the common difference to 4^{th }term. Thus, 5^{th }term of this A.P. will be 15 + 4 = 19.
What would we do if we are asked to find the 20^{th }term or 100^{th }term or n^{th }term? Obviously, the process of adding common difference will be very time consuming.
For such problems, we must have a short cut or a formula to find the general term of an A.P.
Let us derive the same.
Consider the A.P. a, a + d, a + 2d, a + 3d, ...
For this A.P., we have
a_{1} = a
a_{2} – a_{1} = d
a_{3} – a_{2} = d
. . .
. . .
. . .
a_{n }_{– 1} – a_{n }_{– 2} = d
a_{n} – a_{n }_{– 1} = d
Adding all these equations, we get
a_{1} + (a_{2} – a_{1}) + (a_{3} – a_{2}) +...+ (a_{n} – 1 – a_{n }_{– 2}) + (a_{n} – a_{n }_{– 1}) = a + {d + d +... + d (n – 1 times)}
⇒ (a_{1} – a_{1}) + (a_{2 }– a_{2}) + (a_{3} – a_{3}) +...+ (a_{n }_{– 1} – a_{n }_{– 1}) + a_{n} = a + (n – 1)d
⇒ a_{n} = a + (n – 1)d
Hence, the general term or n^{th }term i.e., a_{n} of an A.P. whose first term is a and common difference is d can be found by the following formula:
a_{n} = a + (n – 1)d
Sometimes, we need to find three, four or five consecutive terms of an A.P. then it is convenient to take them as follows:
Three consecutive terms can be taken as a – d, a, a + d
Four consecutive terms can be taken as a – 3d, a – d, a + d, a + 3d. Here, common difference is 2d.
Five consecutive terms can be taken as a – 2d, a – d, a, a + d, a + 2d
Result:
In an A.P., common difference, d = (T_{p} – T_{q})(p – q) , where T_{p} and T_{q} are p^{th }and q^{th }term respectively.
In particular, d = (T_{n} – a)/(n – 1)
Proof:
T_{p} = a + (p – 1)d, T_{q} = a + (q – 1)d
⇒ T_{p} – T_{q} = a + (p – 1)d – {a + (q – 1)d} = (p – q)d
⇒ d = (T_{p} – T_{q})/(p – q)
In particular, take T_{1} = a.
Then, d = (T_{n} – a)/(n – 1)
Now, let us solve some examples to understand the concept better.
Example 1: Find the 20^{th }term of the following arithmetic progression. 0.4, 1.5, 2.6, 3.7, 4.8 …
Answer
Here, a = 0.4 and d = 1.5 – 0.4 = 1.1
Thus, the 20^{th }term is given by,
a_{20} = a + (20 – 1) d
= 0.4 + (20 – 1) 1.1
= 0.4 + 19 × 1.1
= 0.4 + 20.9
= 21.3
Thus, the 20^{th }term of the given A.P. is 21.3.
Example 2: If the 7^{th }term of an A.P. is – 21 and 15^{th }term is – 53, then find the first term and common difference.
Answer
Let the first term and common difference of the A.P. be a and d respectively. It is given that a_{7} = –21 and a_{15} = –53
Using the formula for n^{th }term, we obtain
a_{7} = a + (7 – 1) d
– 21 = a + 6d … (1)
and, a_{15} = a + (15 – 1) d
– 53 = a + 14d … (2)
Subtracting equation (1) from (2), we obtain
–32 = 8d
⇒ d = –4
Substituting the value of d in equation (1), we obtain
–21 = a + 6 (–4)
–21 = a – 24
⇒ a = 3
Thus, the first term is 3 and the common difference is –4.
Example 3: Is 102 a term of the A.P., 5, 11, 17, 23 …?
Answer
Let 102 be the n^{th }term of the given sequence.
∴ a_{n} = 102
Using the formula for n^{th }term, we obtain
a_{n} = a + (n – 1) d
∴ 102 = a + (n – 1) d
For the given A.P., a = 5 and d = 11 – 5 = 6
∴ 102 = 5 + (n – 1) 6
⇒ 102 – 5 = (n – 1) 6
⇒ 97 = (n – 1) 6
⇒ n – 1 = 97/6
⇒ n = 97/6 + 1
⇒ n = 103/6
However, n should be a positive integer. Therefore, 102 is not a term of the given A.P.
Example 4: Find the number of threedigit numbers that are divisible by 5.
Answer
The first threedigit number which is divisible by 5 is 100, second is 105, third is 110, and so on. The last threedigit number which is divisible by 5 is 995.
Thus, we obtain the following A.P.
100, 105 … 995
Here, we have to find the number of terms, n.
∴ Last term of A.P. = 995
The number of terms in the A.P. is n, so the last term is the n^{th }term.
a + (n – 1)d = 995
Here, a = 100 and d = 5
100 + (n – 1)5 = 995
⇒ (n – 1)5 = 995 – 100
⇒ 5n – 5 = 895
⇒ 5n = 895 + 5
⇒ 5n = 900
⇒ n = 900/5
n = 180
Thus, there are 180 threedigit numbers, which are divisible by 5.
Example 5: The fare of a bus is Rs 10 for the first kilometre and Rs 5 for each additional kilometre. Find the fair after 12 kilometres.
Answer
The fare after each kilometre forms an A.P. as follows.
Fare after one kilometre = Rs 10
Fare after two kilometres = 10 + 5 = Rs 15
Fare after three kilometres = 15 + 5 = Rs 20
Now the arithmetic progression is 10, 15, 20 …
Here, first term, a = 10 and common difference, d = 5
Now the fare after 12 kilometres is the 12^{th }term of the A.P.
∴ a_{12} = a + (12 – 1) d
a_{12 }= 10 + 11 × 5
= 10 + 55
= 65
Thus, the fare after 12 kilometres is Rs 65.
Example 6: Mohit borrowed a sum of money at a simple interest rate of 2% per annum. He has to pay an amount of Rs 1120 after 6 years. How much money did he borrow?
Answer
Let the amount of money Mohit borrowed be Rs x. We know that the amount after T years is
A = P + (P × R × T)/100
Where, P and R denotes the principal and rate respectively
The amount after every year forms an A.P.
Amount after first year = x + (x × 2 × 1)/100
= x + 2x/100
Amount after second year = x + (x × 2 × 2)/100
= x + 4x/100
Thus, the A.P. is as follows.
x + 2x/100, x + 4x/100 …….
Here, the first term is x + 2x/100 and common difference is 2x/100 .
Now, it is given that the amount after 6 years is Rs 1120 i.e., 6^{th }term of the A.P. is 1120.
Now using the formula, a_{n} = a + (n – 1)d, we obtain
1120 = x + 2x/100 + (6 – 1)2x/100
⇒ 1120 = x + 2x/100 + 5 × 2x/100
⇒ 1120 = x + 2x/100 + 10x/100
⇒ 1120 = (100x + 2x + 10x)/100
⇒ 1120 × 100 = 112x
⇒ x = 1000
Thus, Mohit borrowed Rs 1000.
Example 7: Find four consecutive terms of an A.P. such that the difference of the middle terms is 8 and the product of the extreme terms is 217.
Answer
Let four consecutive terms of required A.P. be a – 3d, a – d, a + d, a + 3d.
According to the question, we have
a + d – (a – d) = 8
⇒ a + d – a + d = 8
⇒ 2d = 8
⇒ d = 4
Also,
(a – 3d) (a + 3d) = 217
⇒ a^{2 }– (3d)^{2 }= 217
⇒ a^{2 }– 9d^{2 }= 217
⇒ a^{2 }– 9(4^{2}) = 217
⇒ a^{2 }– 144 = 217
⇒ a^{2 }= 361
⇒ a = ±19
When d = 4 and a = 19, then four consecutive terms are:
19 – 3(4), (19 – 4), (19 + 4), 19 + 3(4)
i.e., 7, 15, 23, 31
When d = 4 and a = –19, then four consecutive terms are:
–19 – 3(4), (–19 – 4), (–19 + 4), –19 + 3(4)
i.e., –31, –23, –15, –7
Example 8: Find five consecutive terms of an A.P. such that the product of the extreme terms is –63 and product of second and fourth terms is –15.
Answer
Let five consecutive terms of required A.P. be a – 2d, a – d, a, a + d, a + 2d.
According to the question, we have
(a – 2d)(a + 2d) = –63
⇒ a^{2 }– (2d)^{2 }= – 63
⇒ a^{2 }– 4d^{2 }= – 63 ...(1)
Also,
(a – d)(a + d) = –15
⇒ a^{2 }– d^{2 }= – 15 ...(2)
On subtracting (1) from (2), we get
3d^{2 }= 48
⇒ d^{2 }= 16
⇒ d = ±4
On substituting d^{2 }= 16 in (2), we get
a^{2 }– 16 = –15
⇒ a^{2 }= 1
⇒ a = ±1
When a = 1 and d = 4, A.P. will be –7, –3, 1, 5, 9
When a = 1 and d = –4, A.P. will be 9, 5, 1, –3, –7
When a = –1 and d = 4, A.P. will be –9, –5, –1, 3, 7
When a = –1 and d = –4, A.P. will be 7, 3, –1, –5, –9
Example 9: If the 25^{th }and 35^{th }terms of an arithmetic progression are 121 and 171 respectively, then find the common difference of the A.P.
Answer
T_{25} = 121, T_{35} = 171
It is known that common difference, d = (T_{p} – T_{q})/(p – q).
∴ d = (T_{35} – T_{25})/(35 – 25) = (171 – 121)/10 = 5
Example 10: In an A.P, show that d = (T_{20} – T_{100})/100.
Answer
T_{100 }= a + (100 – 1)d = a + 99d
T_{200} = a + (200 – 1)d = a + 199d
= a + 99d + 100d
= T_{100 }+ 100d
Therefore, d = (T_{200 }– T_{100})/100
Sum of n Terms of an Arithmetic Progression
We know what an arithmetic progression (A.P.) is. Sometimes, we may come across the situations when we have to find the sum of all terms involved in a series and if the series is an AP, then there is a formula which can make the process very simple.
Let us consider a similar situation.
Harry saved Rs 2000 from his salary in the first month. He increases his savings by Rs 50 every month.
Can we calculate his total savings for the first four months?
Let us try to find it.
To find the total savings for the first 4 months, we have to take the sum of the savings for the first four months.
It is given that, savings of Harry for the first month = Rs 2000
Every month, he increases his savings by Rs 50.
Thus, savings for second month = Rs (2000 + 50) = Rs 2050
Similarly, savings for the third month = Rs (2050 + 50) = Rs 2100
nd, savings for the fourth month = Rs (2100 + 50) = Rs 2150
Thus, the total savings of Harry for the first four months = Rs (2000 + 2050 + 2100 + 2150)
= Rs 8300
Now, can we calculate the total savings of Harry for 2 years?
Yes, we can find it as above but it is a very lengthy as well as time consuming process as we have to find the savings for 24 months.
We can also find the total savings of Harry for first two years using a formula. Now, let us see how we can find it.
The savings of Harry for each month forms an A.P., which is as follows. 2000, 2050, 2100, 2150 ….
The sum of savings of Harry = Rs (2000 + 2050 + 2100 + 2150 ….)
Here, we can observe that the total savings of Harry for the first month is the first term of the A.P., i.e. Rs 2000. The total savings for the first two months is the sum of first two terms of the A.P., i.e. Rs (2000 + 2050). In the same way, the total savings of Harry for first 2 years, i.e. 24 months, is the sum of first 24 terms of the A.P. We can find it by using the formula for finding the sum of n terms of an A.P.
Now, let us find the sum of first 24 terms of the above discussed A.P. which is as follows:
2000, 2050, 2100, 2150 ….
Here, first term, a = 2000
Common difference, d = 2050 − 2000 = 50
The sum of first 24 terms of the A.P. is
S_{24} = 24/2[2 × 2000 + (24 – 1)50]
= 12(4000 + 23 × 50)
= 12(4000 + 1150)
= 12 × 5150
= 61800
Therefore, the total savings of Harry for the first two years is Rs 61800.
This formula is used when we are given the first term and the common difference of the arithmetic progression.
We can also find the sum of n terms of an A.P., if we know the first and the last term.
The sum of n terms of an A.P. whose first term is a and last term is l is given by the formula:
S_{n} = n/2{a + 1}
For example, consider an A.P. whose first term is 2 and 30^{th }term is 263. Then, what will be the sum of 30 terms?
Here, a = 2, l = 263 and n = 30
Therefore, sum of 30 terms = n/2[a + l]
= 30/2[2 + 263]
= 15 × 265
= 3975
Thus, the sum of 30 terms is 3975.
Result: The sum of the first n natural numbers is given by n(n + 1)/2
Proof: This can be proved by two methods:
1^{st }method (Using concept of A.P.):
The first n natural numbers can be listed as 1, 2, 3, …, n.
Here, a = 1, d = 1.
∴ S_{n} = n/2[2a + (n – 1)d] = n/2[2 × 1 + (n – 1)1] = n(n + 1)/2
2^{nd }method (Without using concept of A.P.):
S_{n} = 1 + 2 + 3 + … + n … (1)
S_{n} = n + (n – 1) + (n – 2) + … + 3 + 2 + 1 … (2)
Adding (1) and (2):
2S_{n} = (n + 1) + (n + 1) + …… + (n + 1)
Here, there are n terms in the RHS.
∴ S_{n} = n(n+1)/2
Now, let us discuss some more examples based on sum of n terms of an A.P.
Example 1: Find the sum of first 25 terms of the following A.P. 2, 7, 12 …
Answer
Here, a = 2 and d = 7 − 2 = 5.
Sum of the first 25 terms is given by
S_{n }= 25/2 [2a + (25 – 1)d]
= 25/2[2 × 2 + (25 – 1)5]
= 25/2[4 + 24 × 5]
= 25/2[4 + 120]
= 25/2[124]
= 25 × 62
= 1550
Thus, the sum of first 25 terms of the given A.P. is 1550.
Example 2: Find the sum of first 8 terms of the A.P whose n^{th }term is given by 6n − 5.
Answer
The n^{th }term is given by
a_{n} = 6n − 5
On replacing n by 1, 2, 3 … respectively, we get the first, second, third … terms of the A.P.
∴ a_{1} = 6(1) − 5 = 1
a_{2} = 6(2) − 5 = 7
a_{3} = 6(3) − 5 = 13 … and so on.
The A.P. so obtained is as follows.
1, 7, 13 …
Here, the first term, a = 1
and the common difference d = a_{2} − a_{1} = 7 − 1 = 6
Using the formula, S_{n} = n/2[2a + (n – 1)d], the sum of first 8 terms is given by
S_{8} = 8/2[2 × 1 + (8 – 1)6]
= 4[2 + 7 × 6]
= 4[2 + 42]
= 4[44]
= 176
Thus, the sum of first 8 terms is 176.
Example 3: How many terms of the A.P. −28, −24, −20 … should be taken so that the sum will be zero?
Answer
Let the sum of n terms be zero.
Here, a = − 28 and d = − 24 − (−28) = 4
Sum of n terms of an A.P. is given by
S_{n} = n/2[2a + (n – 1)d
But, it is given that the sum of n terms of the given A.P. is zero.
S_{n} = 0
n/2[2a + (n – 1)d] = 0
n/2[2 × (28) + (n – 1) × 4] = 0
=  56 + 4n – 4 = 0
⇒ 4n  60 = 0
⇒ 4n = 60
⇒ n = 60/4
⇒ n = 15
Thus, the sum of 15 terms of the A.P is zero.
Example 4: Sapna’s father planted 4 trees in his garden, when he was 22 years old. After that, every year he planted one more tree than the number of trees he planted in the previous year. How many trees will be there in his garden when he will become 40 years old?
Answer
We can write the given information in the form of an A.P. as follows
Number of trees he planted in the first year = 4
Number of trees he planted in the second year = 4 + 1 = 5
Number of trees he planted in the third year = 5 + 1 = 6
And so on.
Now, the A.P. is 4, 5, 6 …
He planted trees from the age of 22 years to 40 years, i.e. for 19 years.
Thus, we have to find the sum of 19 terms of this A.P.
Here, a = 4, d = 1
And, n = 19
Using the formula, Sn = n/2[2a + (n – 1)d], the sum of 19 terms of this A.P. is given by
S_{19} = 19/2 [2 × 4 + (19 – 1)1]
= 19/2 × [8 + 18]
= 19/2 × 26
= 19 × 13
= 247
Thus, there will be 247 trees in his garden when he will become 40 years old.
Example 5: The p^{th} term of an A.P. is q^{th} and the term of the A.P. is p. What is the sum of (p + q) terms of the A.P.?
Answer
Let the first term and the common difference of the A.P. be a and d respectively.
It is given that the p^{th} term is q.
∴ a_{p} = q
⇒ a + (p − 1) d = q … (1)
Similarly, the q^{th }term is p, therefore we obtain
a + (q − 1) d = p … (2)
On subtracting equation (1) from (2), we obtain
p − q = (q − 1) d − (p − 1) d
⇒ p − q = d [q − 1 − p + 1]
⇒ d = (p – q)/(q – p)
⇒ d =  1
By putting the value of d in equation (1), we obtain
q = a + (p − 1) (−1)
⇒ q = a + 1 − p
⇒ a = p + q − 1
But we know that the sum of n terms of an A.P. is
S_{n} = n/2[2a + (n – 1)d]
Thus, the sum of (p + q) terms is
S_{p+q} = (p + q)/2[2(p + q  1] + (p + q – 1)(1)]
⇒ S_{p+q} = (p + q)/2 [2p + 2q – 2 – p – q + 1]
⇒ S_{p+q }= (p + q)/2[p + q  1]
⇒ S_{p+q} = (p + q)(p + q – 1)/2
Thus, the sum of (p + q) terms is {(p + q)(p + q  1)}/2.
Example 6: Find the n^{th }term of the A.P., the sum of whose n terms is n^{2 }+ 2n.
Answer
Let S_{n} be the sum of n terms.
It is given that the sum of n terms of the A.P. is n^{2 }+ 2n.
∴ S_{n} = n^{2 }+ 2n … (1)
On replacing n by (n − 1) in the equation, we obtain
S_{n – 1} = (n – 1)^{2} + 2(n – 1)
Let a_{n} be the n^{th }term of the A.P.
Therefore, we can write
S_{n} = S_{n }_{−1} + a_{n}
Thus, a_{n} = S_{n }– S_{n}_{−1}
= n^{2 }+ 2n − [(n − 1)^{2 }+ 2(n − 1)]
= n^{2 }+ 2n − [n^{2 }+ 1 − 2n + 2n − 2]
= n^{2 }+ 2n − n^{2 }+ 1
= 2n + 1
Thus, the n^{th }term of the A.P is (2n + 1).
Example 7: Find the sum of first 1000 natural numbers.
Answer
The sum of first n natural number is given by S_{n} = n(n + 1)/2
∴ S_{1000 }= {1000 × (1000 + 1)}/2 = 500 × 1001 = 500500
Hence, the sum of first 1000 natural numbers is 500500.
Example 8: If sum of the first n natural numbers is 5050, find the value of n.
Solution
The sum of first n natural number is given by S_{n} = n(n + 1)/2.
Now,
5050 = {n × (n + 1)}/2
n^{2} + n – 2 × 5050 = 0
⇒ n^{2} – 100n + 101n – 2 × 5050 = 0
⇒ (n – 100)(n + 101)
⇒ n = 100
Hence, n = 100
Properties of Arithmetic Progressions and the Concept of Arithmetic Mean
Arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is a constant. It exhibits some properties which are used in solving various problems.
Properties of an Arithmetic Progression
If a constant is added to each term of an A.P., the resulting sequence will also be an A.P.
If a constant is subtracted from each term of an A.P., the resulting sequence will also be an A.P.
If a constant is multiplied to each term of an A.P., the resulting sequence will also be an A.P.
If each term of an A.P. is divided by a nonzero constant, the resulting sequence is also an A.P.
Arithmetic Mean
If we are given two numbers a and b, then we can insert a number A between these two numbers so that the sequence a, A, b becomes an A.P. Such a number i.e., A is called an arithmetic mean (A.M.) of the numbers a and b.
If A is the A.M. of the numbers a and b, then A is given by A = (a + b)/2.
For example, the A.M. of the two numbers 18 and 16 is (18 + 16)/2 = 17 .
For any two given numbers a and b, we can insert as many numbers between them as we want such that the resulting sequence becomes an A.P.
Solved Examples
Example 1: Between −12 and 40, p numbers have been inserted in such a way that the resulting sequence is an A.P. Find the value of p if the ratio of the 4^{th }and the (p − 3)^{th }number is 1 : 6.
Answer
Let A_{1}, A_{2}, … A_{p} be p numbers such that −12, A_{1}, A_{2}, … A_{p}, 40 is an A.P. Here, a = −12, b = 40, n = p + 2
∴ 40 = −12 + (p + 2 − 1) (d)
⇒ 52 = (p + 1) d
⇒ d = 52/(p + 1) …….(1)
A_{1} = a + d
A_{2 }= a + 2d
A_{3} = a + 3d …
∴ A_{4} = a + 4d
A_{p}_{−3} = a + (p − 3) d
According to the given information,
(a + 4d)/a + (p – 3)d = 1/6
⇒ { 12 + 4(52/(p + 1)}/{ 12 + (p – 3)(52/(p + 1)} = 1/6 [From (1)]
⇒ {12(p + 1) + 4(52)}/{(12(p + 1) + 52(p – 3)} = 1/6
⇒ (12p – 12 + 208)/(12p – 12 + 52p – 156) = 1/6
⇒ (12p + 196)/(40p – 168) = 1/6
⇒ 6(12p + 196) = 40p – 168
⇒ 72p + 1176 = 40p – 168
⇒ 40p – 168 + 72p – 1176 = 0
⇒ 112p = 1344
⇒ p = 1344/112
⇒ p = 12
Thus, the value of p is 12.
Example 2: Between 10 and 30, m numbers are inserted such that the resulting sequence is an A.P. If the sum of all the terms of the A.P. is 140, then find the value of m.
Answer
It is given that between 10 and 30, m numbers are inserted such that the resulting sequence is an A.P.
It is also given that the sum of all the terms in the A.P. is 140.
We know that the sum of n terms of an A.P. is given by
S_{n} = n/2[a + 1], where a is the first term and l is the last term
Here, a = 10, l = 30
Therefore,
140 = n/2[10 + 30]
⇒ 140 = n/2[40]
⇒ 140 = 20n
⇒ n = 7
Thus, the total number of terms in the A.P. is 7.
Thus, the value of m is 7 − 2 = 5.
Geometric Progressions
A sequence a_{1}, a_{2}, …, a_{n}, … is called a geometric progression (G.P.), if each term is nonzero and (a_{k +1})/a_{k} (constant), for k ≥ 1.
For example, 4/7, 4/21, 4/63, 4/189, ……. is in G.P.
A G.P. is written in its standard form as a, ar, ar^{2}, ar^{3}, ar^{4 }…
Here, a is called the first term of the G.P.
Here, r is called the common ratio of the G.P.
There are two types of geometric progressions: finite and infinite.
A finite geometric progression has finite number of terms. In general, a finite G.P. with n terms can be written as a, ar, ar^{2}, ar^{3}, ar^{4 }…ar^{n}^{−1}.
An infinite geometric progression has infinite number of terms. In general, an infinite G.P. can be written as a, ar, ar^{2}, ar^{3}, ar^{4 }… ar^{n}^{−1} ….
There are two types of geometric series: finite and infinite.
The series a + ar + ar^{2 }+ ar^{3 }+ ar^{4 }+… + ar^{n}^{−1 }is a finite geometric series.
The series a + ar + ar^{2 }+ ar^{3 }+ ar^{4 }+… + ar^{n}^{−1 }+ … is an infinite geometric series.
The last term of a G.P. is denoted by l and the sum of n terms of a G.P. is denoted by S_{n}. For example, in the G.P. 8, −16, 32, −64, the last term is given by l = −64.
Note: n^{th} term, a_{n }is also denoted by T_{n}.
Example: In the G.P. 8, 12, 18, 27, 81/2, 243/4,
First term, a = 8
Common ratio, r = 12/8 = 3/2
Last term, l = 243/4
S_{n} = S_{6} = 8 + 12 + 18 + 27 + 81/2 + 243/4
Note:
Since the division by zero is undefined, neither any term nor the common ratio of a G. P. can be 0.
If the common ratio (r) is positive, then all the terms of the G. P. will be of same sign. Thus, all the terms of the G. P. will be either negative or positive.
If the common ratio (r) is negative, then any two consecutive terms of the G. P. will be of different sign. Thus, the G. P. will be having negative and positive terms alternatively.
Solved Examples
Example 1: Which of the following sequences is not a geometric progression?
1. 11, 2.2, 0.44, 0.088, 0.0176 …
2. 17, 17.5, 18, 18.5, 19
3. 81, 81/2, 243/4, 243/8
4. 16, 4, 1, 1/4, 1/16, 1/64 …….
Answer
 The series 11, 2.2, 0.44, 0.088, 0.0176 … is a G.P. since
2.2/11 = 0.44/2.2 = 0.088/0.44 = 0.0176/0.088 = 0.2  The series 17, 17.5, 18, 18.5, 19 is not a G.P. since
17.5/17 ≠ 18/17.5 ≠ 18.5/18 ≠ 19/18.5  The series 81, 81/2, 243/4, 243/8 is not a G.P. since (81/2)/81 = 1/2 and (243/4)/(81/2) = 3/2
 The series 16, 4, 1, 1/4, 1/16, 1/64 ….. is a G.P. since
4/16 = 1/4 = (1/4)/1 = (1/16)/(1/4) = (1/64)/(1/16) = 1/4
Example 2: Write the first term, last term and the common ratio of the following geometric progressions:
1. 3, 6√6, 72, 144√6, 1728
2. 9/5, 18/25, 36/125, 72/625
Answer
1. In the G.P. 3, 6√6, 72, 144√6, 1728
First term, a = 3
Last term, l = 1728
Common Ratio, r = (6√6)/3 = 2√6
2. In the G.P., 9/5, 18/25, 36/125, 72/625,
First term, a = 9/5
Last term, l = 72/625
Common Ratio, r = (18/25)/(9/5) = 18/25 × 5/9 = 2/5
n^{th }Term and Sum of nTerms of a Geometric Progression
The n^{th }term of a geometric progression (G.P.), a, ar, ar^{2}, ar^{3}, … is given by an = ar^{n}^{−1}.
For example, the 20^{th }term of a geometric progression 3, 9/2, 27/4, 81/8 is given by
a_{20} = (3) × (3/2)^{201}
= (3) × (3/2)^{19}
= (3)^{20}/(2)^{19}
Note:
1. To obtain the succeeding term of a given term T_{n}, multiply it by r. T_{n+1} = T_{n} × r
2. To obtain the preceding term of the given term T_{n}, divide it by r.
T_{n  1} = T_{n}/r
If a is the first term and r is the common ratio of a G.P., then the sum of first n terms of the G.P. is given by
S_{n} = na, if r = 1
S_{n} = a(1 – r^{n})/(1  r) or a(r^{n} – 1)/(r – 1) if r ≠ 1
Sum of infinite terms of a G.P.: S_{∞} = a/(1 – r)
Derivation:
S_{n} = a(1 – r^{n})/(1 – r) when r < 1.
Since r^{n }approaches to 0 as n approaches ∞, we have
∴ S_{∞ }= a(1 – 0)/(1 – r) = a/(1 – r)
Note: S_{2n} ÷ S_{n} = r^{n }+ 1
Derivation:
S_{2n} ÷ S_{n} = a(1 – r^{2n})/(1 – r) ÷ a(1 – r^{n})/(1 – r)
= (1 – r^{2n})/(1 – r^{n})
= {(1 – r^{n})(1 + r^{n})}/(1  r^{n})
= 1 + r^{n }
Sometimes, we need to find three or four consecutive terms of an G.P. then it is convenient to take them as follows:
Three consecutive terms can be taken as ar, a, arar, a, ar. Here, common ratio is r.
Four consecutive terms can be taken as ar3, ar, ar, ar3ar3, ar, ar, ar3. Here, common ratio is r2r2.
Solved Examples
Example 1: A scientist kept a solution on fire. The original temperature of the solution was 24°C. He noticed that the temperature of the solution increased by 20% of the original temperature every hour. Find the temperature of the solution after the 5^{th }hour.
Answer
It is given that the original temperature of the solution was 24°C.
It is also given that the temperature of the solution increased by 20% of the original temperature every hour. Hence, the temperature after the 1^{st }hour will be
T_{1} = 20/100 × 24 + 24 = 24(9120/100)
Temperature after 2^{nd }hour:
T_{2} = 24(120/100)(120/100)
Temperature after 3^{rd }hour:
T_{3} = 24(120/100)(120/100)(120/100)
And so on…
Hence, the temperature of the solution noted after every hour will form a geometric progression with
First term, a = 24
Common ratio, r = (120/100) = 1.2
We know that the n^{th }term of a G.P. is given by an = ar^{n}^{−1}.
Also, the temperature of the solution after the 5^{th }hour will be the 6^{th }term of the G.P., which is given by
a_{6} = 24 × (1.2)^{5 }= 59.72 (approx.)
Thus, the temperature of the solution after the 5^{th }hour will be 59.72°C approximately.
Example 2: Find the sum of the sequence 0.4, 0.44, 0.444……. up to n terms.
Answer
The given sequence is not a G.P. However, we can relate it to a G.P. by writing the terms as
S_{n} = 0.4 + 0.44 + 0.444 + … to n terms
= 4[0.1 + 0.11 + 0.111 + ....n terms]
= 4/9[0.9 + 0.99 + 0.999 + ......n terms]
= 4/9[(1 – 0.1) + [1 –(0.1)^{2}] + [1 – (0.1)^{3}] + ..... n terms]
= 4/9 {(1 + 1 + 1 .....n terms) – [0.1 + (0.1)^{2} + (0.1)^{3} + .... n terms]}
= 4/9[n – 0.1 × (1 – (0.1)^{n})/(1 – 0.1)
Example 3: Find the number of terms of the G.P. 6, 12/5, (24/25), 48/125, (96/625) that are required for giving the sum as – (2706/625).
Answer
Let n be the required number of terms.
It is given that
First term, a = −6
Common ratio, r = (2/5)
We know that the sum of n terms of a G.P. is given by Sn = a(1 – r^{n})/(1 – r).
Therefore,
S_{n} = (6)[1 – (2/5)^{n}]/[(1 – (2/5)]
⇒ 2706/625 = (6)[1 – (2/5)^{n}]/(7/5)
⇒ [1 – (2/5)^{n}] = 2706/625 × 7/5 × 1/6 = 3157/3125
⇒ (2/5)^{n} = 1 – 3157/3125 = 32/3125
⇒ (2/5)^{n} = (2/5)^{5}
⇒ n = 5
Thus, the required number of terms of the G.P. is 5.
Example 4: In a G.P., the ratio of the 11^{th }term and the 14^{th }term is given by 27:125. Find the 18^{th }term of the G.P. if the 5^{th }term of the G.P. is 125/81.
Answer
Let the first term and the common ratio of the G.P. be a and r respectively.
We know that the n^{th }term of a G.P. is given by an = ar^{n}^{−1}. Therefore, a_{11} = ar^{10}
a_{14 }= ar^{13}
It is given that the ratio of the 11^{th }term to the 14^{th }term is 27:125. Hence,
ar^{10}/ar^{13} = 27/125
⇒ 1/r^{3} = (3/5)^{3}
⇒ 1/r = 3/5
⇒ r = 5/3
It is also given that the 5^{th }term of the G.P. is 125/81. Hence,
a_{5} = ar^{4} = 125/81
⇒ a = (5/3)^{4} = 125/81
⇒ a × 625/81 = 125/81
⇒ a = 1/5
Thus, the 18^{th }term of the G.P. is given by
a_{18} = ar_{17} = (1/5)(5/3)17 = (5)^{16}/(3)^{17}
Example 5: If S_{n} denotes the sum of n terms of a geometric series such that S_{10}/S_{5} = 33/32, find the value of the common ratio.
Answer
S_{10}/S_{5} = 33/32
S_{(2×5)}/S_{5} = 33/32
r^{5} + 1 = 33/32 [S_{2n}/S_{n} = r^{n} + 1]
⇒ r^{5} + 1 = (1/2)^{5} + 1
⇒ r = 1/2
Thus, the value of the common ratio is 1/2.
Example 6: For a G.P. with the first term as 2, if S_{∞} = 5/2, then find S_{3}.
Answer
S_{∞ }= a/(1 – r)
⇒ 5/2 = 2/(1 – r)
⇒ 2/5 = (1 – r)/2
⇒ 4/5 = 1 – r
⇒ r = 1 – 4/5 = 1/5
⇒ S_{n} = a(1 – r^{n})/(1 – r) = 2(1 – (1/5)^{3})/(1 – 1/5) = 2(124/125)/(4/5)
= 2(124/125) × 5/4 = 62/25
Example 7: The sum of three consecutive terms of a geometric progression is 21/2 and their product is 27. Find the terms.
Answer
Let the three consecutive terms of G.P. be a/r, a, ar. Then,
a/r + a + ar = 21/1
⇒ a_{1}/r + 1 + r) = 21/2
⇒ a(1 + r + r2)/r = 21/2 ....(i)
Also, a/r × a × ar = 27
⇒ a^{3} = 27
⇒ a = 3
Substituting the value of a in (i),
a(1 + r + r^{2}/r) = 21/2
⇒ 3(1 + r + r2/r) = 21/2
⇒ (1 + r + r2)/r = 7/12
⇒ 2 + 2r + 2r2 = 7r
⇒ 2r^{2} – 5r + 2 = 0
⇒ 2r^{2} – 4r – r + 2 = 0
⇒ 2r(r – 2) – 1(r – 2) = 0
⇒ (r – 2)(2r – 1) = 0
⇒ r = 2 or r = 1/2
So, the terms are 3/2, 3, 6 or 6, 3, 3/2
Geometric Mean and Its Relation with Arithmetic Mean
Like arithmetic mean, geometric mean (G. M.) is also a number which lies between two different numbers.
Arithmetic mean is the number which gives the common difference with the numbers on its left and right.
For example, if a, b, c is a sequence of numbers such that b is the arithmetic mean of a and c then we have b – a = c – b.
Similarly, geometric mean is the number which gives the common ratio with the numbers on its left and right.
For example, if a, b, c is a sequence of numbers such that b is the geometric mean of a and c, then we have
b/a = c/b
⇒ b^{2} = ac
⇒ b = √ac
Hence the geometric mean (G.M.) of two numbers a and b is and it is denoted by G.
For example, the G.M. of the two numbers 36 and 25 is = 30.
Note:
G comes out be a real number if and only if both a and b have the same sign.
For any two numbers a and b, there exist two values of G such that
If a = b then GM(G) = ±a and AM (A) = (a + a)/2 = a.
If we are given two numbers a and b, then we can insert a number G between these two numbers so that the sequence a, G, b becomes a G.P. Here, G is the geometric mean (G.M.) of the numbers a and b.
For any two positive numbers, we can insert as many numbers between them as we want so that the resulting sequence becomes a G.P.
Now, let us understand this concept with the help of geometry.
Consider a line segment MP = a units and PN = b units as shown in the following figure.
Thus, MN = MP + PN = (a + b) units
It can be seen that that a semicircle is drawn with diameter MN and OP is perpendicular to MN such that point O lies on the circle. On joining O to M and N, we get two right angled triangles such as Î”MPO and Î”NPO.
Since Î”MON is inscribed in a semicircle, it is also a rightangled triangle such that ∠MON = 90°.
Example 6: For a G.P. with the first term as 2, if S_{∞} = 5/2, then find S_{3}.
Answer
S_{∞ }= a/(1 – r)
⇒ 5/2 = 2/(1 – r)
⇒ 2/5 = (1 – r)/2
⇒ 4/5 = 1 – r
⇒ r = 1 – 4/5 = 1/5
⇒ S_{n} = a(1 – r^{n})/(1 – r) = 2(1 – (1/5)^{3})/(1 – 1/5) = 2(124/125)/(4/5)
= 2(124/125) × 5/4 = 62/25
Example 7: The sum of three consecutive terms of a geometric progression is 21/2 and their product is 27. Find the terms.
Answer
Let the three consecutive terms of G.P. be a/r, a, ar. Then,
a/r + a + ar = 21/1
⇒ a_{1}/r + 1 + r) = 21/2
⇒ a(1 + r + r2)/r = 21/2 ....(i)
Also, a/r × a × ar = 27
⇒ a^{3} = 27
⇒ a = 3
Substituting the value of a in (i),
a(1 + r + r^{2}/r) = 21/2
⇒ 3(1 + r + r^{2}/r) = 21/2
⇒ (1 + r + r^{2})/r = 7/12
⇒ 2 + 2r + 2r^{2} = 7r
⇒ 2r^{2} – 5r + 2 = 0
⇒ 2r^{2} – 4r – r + 2 = 0
⇒ 2r(r – 2) – 1(r – 2) = 0
⇒ (r – 2)(2r – 1) = 0
⇒ r = 2 or r = 1/2
So, the terms are 3/2, 3, 6 or 6, 3, 3/2
Geometric Mean and Its Relation with Arithmetic Mean
Like arithmetic mean, geometric mean (G. M.) is also a number which lies between two different numbers.
Arithmetic mean is the number which gives the common difference with the numbers on its left and right.
For example, if a, b, c is a sequence of numbers such that b is the arithmetic mean of a and c then we have b – a = c – b.
Similarly, geometric mean is the number which gives the common ratio with the numbers on its left and right.
For example, if a, b, c is a sequence of numbers such that b is the geometric mean of a and c, then we have
Hence the geometric mean (G.M.) of two numbers a and b isand it is denoted by G.
For example, the G.M. of the two numbers 36 and 25 is
Note:
G comes out be a real number if and only if both a and b have the same sign.
For any two numbers a and b, there exist two values of G such that
If we are given two numbers a and b, then we can insert a number G between these two numbers so that the sequence a, G, b becomes a G.P. Here, G is the geometric mean (G.M.) of the numbers a and b.
For any two positive numbers, we can insert as many numbers between them as we want so that the resulting sequence becomes a G.P.
Now, let us understand this concept with the help of geometry.
Consider a line segment MP = a units and PN = b units as shown in the following figure.
Thus, MN = MP + PN = (a + b) units
It can be seen that that a semicircle is drawn with diameter MN and OP is perpendicular to MN such that point O lies on the circle. On joining O to M and N, we get two right angled triangles such as Î”MPO and Î”NPO.
Since Î”MON is inscribed in a semicircle, it is also a rightangled triangle such that ∠MON = 90°.
In the given figure, we have
∠MON = 90°
⇒ ∠MOP + ∠NOP = 90° ...(1)
Also,
∠PON + ∠PNO + ∠OPN = 180° (By angle sum property in Î”OPN)
⇒ ∠PON + ∠PNO + 90° = 180°
⇒ ∠PON + ∠PNO = 90° ...(2)
From (1) and (2), we get
∠MOP = ∠PNO ...(3)
Now, in Î”MPO and Î”OPN,
∠MOP = ∠PNO
And, ∠OPM = ∠OPN
Thus, by AA similarity, these triangles are similar.
Using property of similar triangles, we get
MP/OP = OP/PN
⇒ OP^{2} = MP × PN
⇒ OP^{2} = ab
Thus, OP is the geometric mean of MP and PN.
Hence, it can be concluded that the length of perpendicular drawn from any point on the circumference to the diameter is the geometric mean of the lengths of the two segments obtained on diameter.
Relation between A.M. and G.M.
If A and G are the respective arithmetic mean and geometric mean of the numbers a and b, then we will always have the following relation between A and G:
Solved Examples
Example 1: Insert five numbers between 2/3 and 2/2187 so that the resulting sequence is a G.P.
Answer
Let G_{1}, G_{2}, G_{3}, G_{4} and G_{5} be the five numbers between and such that –(2/3), G_{1}, G_{2}, G_{3}, G_{4}, G_{5}, 2/2187 is a G.P.
We know that the n^{th }term of a G.P. is given by
a_{n} = ar^{n}^{−1}
Therefore,
2/2187 = 2/3 × r^{6}
⇒ r^{6} = 1/729
⇒ r^{6} = (1/3)^{6}
⇒ r = ± 1/3
For r = 1/3,
G_{1} = ar = (2/3) × 1/3 = 2/9
G_{2} = ar^{2 }= (2/3) × (1/3)^{2} =  (2/27)
G_{3} = ar^{3 }= (2/3) × (1/3)^{3} = (2/81)
G_{4} = ar^{4 }= (2/3) × (1/3)^{4} = (2/243)
G_{5} = ar^{5 }= (2/3) × (1/3)^{5 }= (2/729)
For r = (1/3)
G_{1} = ar = (2/3) × (1/3) = 2/9
G_{2} = ar^{2 }= (2/3) × (1/3)^{2} =  (2/27)
G_{3} = ar^{3 }= (2/3) × (1/3)^{3} = (2/81)
G_{4} = ar^{4 }= (2/3) × (1/3)^{4} = (2/243)
G_{5} = ar^{5 }= (2/3) × (1/3)^{5 }= (2/729)
Thus, we can obtain two G.P.s by inserting five numbers between the given numbers –(2/3) and (2/2187). The two G.P.s are –(2/3), (2/9), (2/27), (2/81), (2/243), (2/729), (2/2187) and –(2/3), 2/9, (2/27), 2/81, (2/243), (2/729), 2/2187.
Example 2: The ratio of two numbers is 4:1. If the arithmetic mean of the two numbers is 5 more than their geometric mean, then find the two numbers.
Answer
It is given that the ratio of the two numbers is 4:1. Hence, let the two numbers be 4 x and x.
The arithmetic mean, A, is given by
(4x + x)/2 = 5x/2
The geometric mean, G, is given by
It is given that the arithmetic mean of the two numbers is 5 more than their geometric mean. Therefore,
5x/2 = 2x + 5
⇒ 5x = 4x + 10
⇒ x = 10
Thus, the two numbers are 10 and 4 × 10 = 40