Welcome, Guest  0 
ekShiksha : Privacy Setting
5Arithmetic Progressions
1 Introduction
You must have observed that in nature, many things follow a certain pattern, such as the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirals on a pineapple and on a pine cone etc.
We now look for some patterns which occur in our daytoday life. Some such examples are :
 Reena applied for a job and got selected. She
has been offered a job with a starting monthly
salary of Rs 8000, with an annual increment of
Rs 500 in her salary. Her salary (in Rs) for the
1st, 2nd, 3rd, . . . years will be, respectively
8000, 8500, 9000, . . . .
 The lengths of the rungs of a ladder decrease
uniformly by 2 cm from bottom to top
(see Fig. 5.1). The bottom rung is 45 cm in
length. The lengths (in cm) of the 1st, 2nd,
3rd, . . ., 8th rung from the bottom to the top
are, respectively
45, 43, 41, 39, 37, 35, 33, 31
Fig 1  In a savings scheme, the amount becomes 5/4 times of itself after every 3 years.The maturity amount (in Rs) of an investment of Rs 8000 after 3, 6, 9 and 12 years will be, respectively :
 The number of unit squares in squares with side 1, 2, 3, . . . units (see Fig. 5.2)
are, respectively
1^{2},2^{2},3^{2},.......
Fig. 2  Shakila put Rs 100 into her daughter’s money box when she was one year old
and increased the amount by Rs 50 every year. The amounts of money (in Rs) in
the box on the 1st, 2nd, 3rd, 4th, . . . birthday were
100, 150, 200, 250, . . ., respectively.
 A pair of rabbits are too young to produce in their first month. In the second, and
every subsequent month, they produce a new pair. Each new pair of rabbits
produce a new pair in their second month and in every subsequent month (see
Fig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start of
the 1st, 2nd, 3rd, . . ., 6th month, respectively are :
1, 1, 2, 3, 5, 8
Fig 3
10000, 12500, 15625, 19531.25
In the examples above, we observe some patterns. In some, we find that the succeeding terms are obtained by adding a fixed number, in other by multiplying with a fixed number, in another we find that they are squares of consecutive numbers, and so on.
In this chapter, we shall discuss one of these patterns in which succeeding terms are obtained by adding a fixed number to the preceding terms. We shall also see how to find their nth terms and the sum of n consecutive terms, and use this knowledge in solving some daily life problems.
2 Arithmetic Progressions
Consider the following lists of numbers :
 1, 2, 3, 4, . . .
 100, 70, 40, 10, . . .
 – 3, –2, –1, 0, . . .
 3, 3, 3, 3, . . .
 –1.0, –1.5, –2.0, –2.5, . . .
Each of the numbers in the list is called a term
Given a term, can you write the next term in each of the lists above? If so, how will you write it? Perhaps by following a pattern or rule. Let us observe and write the rule.
In (i), each term is 1 more than the term preceding it.
In (ii), each term is 30 less than the term preceding it.
In (iii), each term is obtained by adding 1 to the term preceding it.
In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding (or subtracting) 0 to the term preceding it
In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the term preceding it.
In all the lists above, we see that successive terms are obtained by adding a fixed number to the preceding terms. Such list of numbers is said to form an Arithmetic Progression ( AP ).
So, an arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
This fixed number is called the common difference of the AP. Remember that it can be positive, negative or zero.
Consider first a linear polynomial ax + b, a ≠ 0. You have studied in Class IX that the graph of y = ax + b is a straight line. For example, the graph of y = 2x + 3 is a straight line passing through the points (– 2, –1) and (2, 7).
Let us denote the first term of an AP by a1, second term by a_{2}, . . ., nth term by an and the common difference by d. Then the AP becomes a_{1}, a_{2}, a_{3}, . . ., a_{n}.
So, a_{2} – a_{1} = a_{3} – a_{2} = . . . = a_{n} – a_{n1} = d.
Some more examples of AP are:
 The heights ( in cm ) of some students of a school standing in a queue in the morning assembly are 147 , 148, 149, . . ., 157.
 The minimum temperatures ( in degree celsius ) recorded for a week in the
month of January in a city, arranged in ascending order are
– 3.1, – 3.0, – 2.9, – 2.8, – 2.7, – 2.6, – 2.5
 The balance money ( in Rs ) after paying 5 % of the total loan of Rs 1000 every month is 950, 900, 850, 800, . . ., 50.
 The cash prizes ( in Rs ) given by a school to the toppers of Classes I to XII are, respectively, 200, 250, 300, 350, . . ., 750.
 The total savings (in Rs) after every month for 10 months when Rs 50 are saved each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
It is left as an exercise for you to explain why each of the lists above is an AP.
You can see that
a, a + d, a + 2d, a + 3d, . . .
represents an arithmetic progression where a is the first term and d the common difference. This is called the general form of an AP.
Note that in examples (a) to (e) above, there are only a finite number of terms. Such an AP is called a finite AP. Also note that each of these Arithmetic Progressions (APs) has a last term. The APs in examples (i) to (v) in this section, are not finite APs and so they are called infinite Arithmetic Progressions. Such APs do not have a last term.
Now, to know about an AP, what is the minimum information that you need? Is it enough to know the first term? Or, is it enough to know only the common difference? You will find that you will need to know both – the first term a and the common difference d.
For instance if the first term a is 6 and the common difference d is 3, then the AP is
6, 9,12, 15, . . .
and if a is 6 and d is – 3, then the AP is
6, 3, 0, –3, . . .
Similarly, when
a = – 7, d = – 2, the AP is – 7, – 9, – 11, – 13, . . .
a = 1.0, d = 0.1, the AP is 1.0, 1.1, 1.2, 1.3, . . .
a = 0, d=1^{1}/_{2}, the AP is 0, 1^{1}/_{2},3,4^{1}/_{2},6,....
a = 2, d = 0, the AP is 2, 2, 2, 2, . . .
So, if you know what a and d are, you can list the AP. What about the other way round? That is, if you are given a list of numbers can you say that it is an AP and then find a and d? Since a is the first term, it can easily be written. We know that in an AP, every succeeding term is obtained by adding d to the preceding term. So, d found by subtracting any term from its succeeding term, i.e., the term which immediately follows it should be same for an AP.
For example, for the list of numbers :
6, 9, 12, 15, . . . ,
We have a_{2} – a_{1} = 9 – 6 = 3,
a_{3} – a_{2} = 12 – 9 = 3,
a_{4} – a_{3} = 15 – 12 = 3
Here the difference of any two consecutive terms in each case is 3. So, the given list is an AP whose first term a is 6 and common difference d is 3.
For the list of numbers : 6, 3, 0, – 3, . . .,
a_{2} – a_{1} = 3 – 6 = 3,
a_{3} – a_{2} = 0 – 3 = 3,
a_{4} – a_{3} = 3 – 0 = 3
Similarly this is also an AP whose first term is 6 and the common difference is –3.
In general, for an AP a_{1}, a_{2}, . . ., an, we have
d = a_{k+1} – a
where a_{k+1} and a_{k} are the ( k + 1)th and the kth terms respectively.
To obtain d in a given AP, we need not find all of a_{2} – a_{1}, a_{3} – a_{2}, a_{4} – a_{3}, . . . . It is enough to find only one of them.
Consider the list of numbers 1, 1, 2, 3, 5, . . . . By looking at it, you can tell that the difference between any two consecutive terms is not the same. So, this is not an AP.
Note that to find d in the AP : 6, 3, 0, – 3, . . ., we have subtracted 6 from 3 and not 3 from 6, i.e., we should subtract the kth term from the (k + 1) th term even if the (k + 1) th term is smaller.
Let us make the concept more clear through some examples.
For the AP : 3/2, 1/2, (1/2), (3/2),...., write the first term a and the common difference d.
Here, a=3/2, d=[1/2  3/2] = 1.
Remember that we can find d using any two consecutive terms, once we know that the numbers are in AP.
Which of the following list of numbers does form an AP? If they form an AP, write the next two terms :
 4, 10, 16, 22, . . .
 1, – 1, – 3, – 5, . . .
 – 2, 2, – 2, 2, – 2, . . .
 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .

We have a_{2} – a_{1} = 10 – 4 = 6
a_{3} – a_{2} = 16 – 10 = 6
a_{4} – a_{3} = 22 – 16 = 6
i.e., a_{k+1} – a_{k} is the same every time.So, the given list of numbers forms an AP with the common difference d = 6. The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.

a_{2} – a_{1} = – 1 – 1 = – 2
a_{3} – a_{2} = – 3 – ( –1 ) = – 3 + 1 = – 2
a_{3} – a_{3} = – 5 – ( –3 ) = – 5 + 3 = – 2
a_{k+1} – a_{k} is the same every time.
So, the given list of numbers forms an AP with the common difference d = – 2.
The next two terms are:
– 5 + (– 2 ) = – 7 and – 7 + (– 2 ) = – 9 
a_{2} – a_{1} = 2 – (– 2) = 2 + 2 = 4
a_{3} – a_{2} = – 2 – 2 = – 4
As a_{2} – a_{1} a_{3} – a_{2} , the given list of numbers does not form an AP.

a_{2} – a_{1} = 1 – 1 = 0
a_{3} – a_{2} = 1 – 1 = 0
a_{4} – a_{3} = 2 – 1 = 1
Here, a_{2} – a_{1} = a_{3} – a_{2} ≠ a_{4} – a_{3}.
So, the given list of numbers does not form an AP.
3 nth Term of an AP
Let us consider the situation again, given in Section 5.1 in which Reena applied for a job and got selected. She has been offered the job with a starting monthly salary of Rs 8000, with an annual increment of Rs 500. What would be her monthly salary for the fifth year?
To answer this, let us first see what her monthly salary for the second year would be.
It would be Rs (8000 + 500) = Rs 8500. In the same way, we can find the monthly salary for the 3rd, 4th and 5th year by adding Rs 500 to the salary of the previous year. So, the salary for the 3rd year = Rs (8500 + 500)
= Rs (8000 + 500 + 500)
= Rs (8000 + 2 × 500)
= Rs [8000 + (3 – 1) × 500] (for the 3rd year)
= Rs 9000
Salary for the 4th year
= Rs (9000 + 500)
= Rs (8000 + 500 + 500 + 500)
= Rs (8000 + 3 × 500)
= Rs [8000 + (4 – 1) × 500] (for the 4th year)
= Rs 9500
Salary for the 5th year = Rs (9500 + 500)
= Rs (8000+500+500+500 + 500)
= Rs (8000 + 4 × 500)
= Rs [8000 + (5 – 1) × 500] (for the 5th year)
= Rs 10000
Observe that we are getting a list of numbers
8000, 8500, 9000, 9500, 10000, . . .
These numbers are in AP. (Why?)
Now, looking at the pattern formed above, can you find her monthly salary for the 6th year? The 15th year? And, assuming that she will still be working in the job, what about the monthly salary for the 25th year? You would calculate this by adding Rs 500 each time to the salary of the previous year to give the answer. Can we make this process shorter? Let us see. You may have already got some idea from the way we have obtained the salaries above.
Salary for the 15th year
= Salary for the 14th year + Rs 500
= Rs [8000 + (500 + 500 + 500 + . . . + 500) _{13 times}] + Rs 500
= Rs [8000 + 14 × 500]
= Rs [8000 + (15 – 1) × 500] = Rs 15000
i.e., First salary + (15 – 1) × Annual increment.
In the same way, her monthly salary for the 25th year would be
Rs [8000 + (25 – 1) × 500] = Rs 20000
= First salary + (25 – 1) × Annual increment
This example would have given you some idea about how to write the 15th term, or the 25th term, and more generally, the nth term of the AP.
Let a_{1} , a_{2} , a_{3} , . . . be an AP whose first term a_{1} is a and the common difference is d.
Then, the second term a_{2} = a + d = a + (2 – 1) d
the third term a3 = a2 + d = (a + d) + d = a + 2d = a + (3 – 1) d
the fourth term a4 = a3 + d = (a + 2d) + d = a + 3d = a + (4 – 1) d
........
........
Looking at the pattern, we can say that the nth term an = a + (n – 1) d.
So, the nth term an of the AP with first term a and common difference d is given by an = a + (n – 1) d.
a_{n} is also called the general term of the AP. If there are m terms in the AP, then a_{m} represents the last term which is sometimes also denoted by l.
Let us consider some examples.
Find the 10th term of the AP : 2, 7, 12, . . .
Here, a = 2, d=7–2=5 and n = 10.
We have
a_{n} = a + (n – 1) d
So, a_{10} = 2 + (10 – 1) × 5 = 2 + 45 = 47
Therefore, the 10th term of the given AP is 47.
Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer.
Here, a = 21, d = 18 – 21 = – 3 and an = – 81, and we have to find n.
As a_{n} = a + ( n – 1) d,
we have – 81 = 21 + (n – 1)(– 3)
– 81 = 24 – 3n
– 105 = – 3n
So, n = 35
Therefore, the 35th term of the given AP is – 81.
Next, we want to know if there is any n for which a_{n} = 0. If such an n is there, then
21 + (n – 1) (–3) = 0,
i.e., 3(n – 1) = 21
i.e., n= 8
So, the eighth term is 0.
Determine the AP whose 3rd term is 5 and the 7th term is 9.
We have
a_{3} = a + (3 – 1) d = a + 2d = 5 (1)
and a_{7} = a + (7 – 1) d = a + 6d = 9 (2)
Solving the pair of linear equations (1) and (2), we get
a = 3, d = 1
Hence, the required AP is 3, 4, 5, 6, 7, . . .
Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . .
We have :
a_{2} – a_{1} = 1_{1} – 5 = 6, a_{3} – a_{2} = 17 – 11 = 6, a_{4} – a_{3} = 23 – 17 = 6
As a_{k+1} – a_{k} is the same for k = 1, 2, 3, etc., the given list of numbers is an AP.
Now, a=5 and d = 6.
Let 301 be a term, say, the nth term of the this AP.
We know that
a_{n} = a + (n – 1) d
So, 301 = 5 + (n – 1) × 6
i.e., 301 = 6n – 1
So, n = 302/6 = 151/3
But n should be a positive integer (Why?). So, 301 is not a term of the given list of numbers.
How many twodigit numbers are divisible by 3?
The list of twodigit numbers divisible by 3 is :
12, 15, 18, . . . , 99
Is this an AP? Yes it is. Here, a = 12, d = 3, an = 99.
As a_{n} = a + (n – 1) d,
we have 99 = 12 + (n – 1) × 3
i.e., 87 = (n – 1) × 3
i.e., n–1 = 87/3 = 29
i.e., n = 29 + 1 = 30
So, there are 30 twodigit numbers divisible by 3.
Find the 11th term from the last term (towards the first term) of the AP : 10, 7, 4, . . ., – 62.
Here, a = 10, d = 7 – 10 = – 3, l = – 62,
where l = a + (n – 1) d
To find the 11th term from the last term, we will find the total number of terms in the AP.
So, – 62 = 10 + (n – 1)(–3)
i.e., – 72 = (n – 1)(–3)
i.e., n – 1 = 24
or n = 25
So, there are 25 terms in the given AP.
The 11th term from the last term will be the 15th term. (Note that it will not be the 14th term. Why?)
So, a_{15} = 10 + (15 – 1)(–3) = 10 – 42 = – 32
i.e., the 11th term from the last term is – 32.
If we write the given AP in the reverse order, then a = – 62 and d = 3 (Why?) So, the question now becomes finding the 11th term with these a and d.
So, a_{11} = – 62 + (11 – 1) × 3 = – 62 + 30 = – 32
So, the 11th term, which is now the required term, is – 32.
A sum of Rs 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.
We know that the formula to calculate simple interest is given by
Simple Interest = (P× R × T) / 100
So, the interest at the end of the 1st year = Rs (1000x8x1) / 100 = Rs 80
The interest at the end of the 2nd year = Rs (1000x8x2) / 100 = Rs 160
The interest at the end of the 3rd year = Rs (1000x8x3) / 100 = Rs 240
Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on.
So, the interest (in Rs) at the end of the 1st, 2nd, 3rd, . . . years, respectively are
80, 160, 240, . . .
It is an AP as the difference between the consecutive terms in the list is 80, i.e., d = 80. Also, a = 80.
So, to find the interest at the end of 30 years, we shall find a_{30}.
Now, a_{30} = a + (30 – 1) d = 80 + 29 × 80 = 2400
So, the interest at the end of 30 years will be Rs 2400.
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are :
23, 21, 19, . . ., 5
It forms an AP (Why?). Let the number of rows in the flower bed be n.
Then a = 23, d = 21 – 23 = – 2, a_{n} = 5
As, a_{n} = a + (n – 1) d
We have, 5 = 23 + (n – 1)(– 2)
i.e., – 18 = (n – 1)(– 2)
i.e., n = 10
So, there are 10 rows in the flower bed.
4Sum of First n Terms of an AP
Let us consider the situation again given in Section 5.1 in which Shakila put Rs 100 into her daughter’s money box when she was one year old, Rs 150 on her second birthday, Rs 200 on her third birthday and will continue in the same way. How much money will be collected in the money box by the time her daughter is 21 years old?
Here, the amount of money (in Rs) put in the money box on her first, second, third, fourth . . . birthday were respectively 100, 150, 200, 250, . . . till her 21st birthday. To find the total amount in the money box on her 21st birthday, we will have to write each of the 21 numbers in the list above and then add them up. Don’t you think it would be a tedious and time consuming process? Can we make the process shorter? This would be possible if we can find a method for getting this sum. Let us see.
We consider the problem given to Gauss (about whom you read in Chapter 1), to solve when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can you guess how did he do? He wrote :
S = 1 + 2 + 3 + . . . + 99 + 100
And then, reversed the numbers to write
S = 100 + 99 + . . . + 3 + 2 + 1
Adding these two, he got
2S = (100 + 1) + (99 + 2) + . . . + (3 + 98) + (2 + 99) + (1 + 100)
= 101 + 101 + . . . + 101 + 101 (100 times)
So, S = (100x101)/2 = 5050 , i.e., the sum = 5050.
We will now use the same technique to find the sum of the first n terms of an AP :
a, a + d, a + 2d, . . .
The nth term of this AP is a + (n – 1) d. Let S denote the sum of the first n terms of the AP. We have
S = a + (a + d ) + (a + 2d ) + . . . + [a + (n – 1) d ] (1)
Rewriting the terms in reverse order, we have
S = [a + (n – 1) d ] + [a + (n – 2) d ] + . . . + (a + d ) + a (2)
On adding (1) and (2), termwise. we get
2S = [2 a + ( n − 1) d ] + [2 a + ( n − 1) d ] + ... + [2 a + ( n − 1) d ] + [2 a + ( n − 1) d ]_{n times} 4
or,
or,
2S = n [2a + (n – 1) d ] (Since, there are n terms)
S = (n/2)[2a + (n – 1) d ]
So, the sum of the first n terms of an AP is given by
S = (n/2)[2a + (n – 1) d ]
We can also write this as S = (n/2)[a + a + (n – 1) d ]
i.e., S = (n/2)[a + a_{n}] (3)
Now, if there are only n terms in an AP, then a_{n} = l, the last term. From (3), we see that
S = (n/2)[a + l] (4)
This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.
Now we return to the question that was posed to us in the beginning. The amount of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday, . . ., were 100, 150, 200, 250, . . ., respectively.
This is an AP. We have to find the total money collected on her 21st birthday, i.e., the sum of the first 21 terms of this AP.
Here, a = 100, d = 50 and n = 21. Using the formula :
S = (n/2)[2a + (n – 1) d ]
we have S= (21/2)[ 2 × 100 + (21 − 1) × 50] = 21/2[200 + 1000]
= 21/2 × 1200 = 12600
So, the amount of money collected on her 21st birthday is Rs 12600.
Hasn’t the use of the formula made it much easier to solve the problem?
We also use S_{n} in place of S to denote the sum of first n terms of the AP. We write S_{20} to denote the sum of the first 20 terms of an AP. The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth.
Remark :
The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., an = S_{n} – S_{n1}.
Let us consider some examples.
Find the sum of the first 22 terms of the AP : 8, 3, –2, . .
Here, a = 8, d = 3 – 8 = –5, n = 22.
We know that
S = (n/2)[2a + (n – 1) d ]
Therefore, S = (22/2)[16 + 21 (−5)] = 11(16 – 105) = 11(–89) = – 979
So, the sum of the first 22 terms of the AP is – 979.
If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Here, S_{14} = 1050, n = 14, a = 10.
As S_{n} = (n/2)[2a + (n – 1) d ]
so, 1050 = (14/2)[20_13d] = 140 + 91d
i.e., 910 = 91d
or, d = 10
Therefore, a_{20} = 10 + (20 – 1) × 10 = 200, i.e. 20th term is 200.
How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?
Here, a = 24, d = 21 – 24 = –3, Sn = 78. We need to find n.
We know that S_{n} = (n/2)[2a + (n – 1) d ]
So, 78 = (n/2)[48 + (n − 1)( −3)] = (n/2)[513n]
or 3n^{2} – 51n + 156 = 0
or n^{2} – 17n + 52 = 0
or (n – 4)(n – 13) = 0
or n = 4 or 13
Both values of n are admissible. So, the number of terms is either 4 or 13.
Remarks :
 In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.
 Two answers are possible because the sum of the terms from 5th to 13th will be zero. This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.
Find the sum of :
 the first 1000 positive integers
 the first n positive integers
 Let S = 1 + 2 + 3 + . . . + 1000
Using the formula S_{n} = (n/2)(a+l) for the sum of the first n terms of an AP, we have
S_{1000} = (1000/2)(1+1000)= 500 × 1001 = 500500
So, the sum of the first 1000 positive integers is 500500.
 Let S_{n} = 1 + 2 + 3 + . . . + n
Here a = 1 and the last term l is n.
Therefore, S_{n} = [n(1 + n)]/2 or S_{n} = [n(n+1)/2]
So, the sum of first n positive integers is given by
S_{n} = [n(n+1)/2]
Find the sum of first 24 terms of the list of numbers whose nth term is given by
a_{n} = 3 + 2n
As a_{n} = 3 + 2n
so, a_{1} = 3 + 2 = 5
a_{2} = 3 + 2 × 2 = 7
a_{3} = 3 + 2 × 3 = 9
List of numbers becomes 5, 7, 9, 11, . . .
Here, 7 – 5 = 9 – 7 = 11 – 9 = 2 and so on.
So, it forms an AP with common difference d = 2.
To find S_{24}, we have n = 24, a = 5, d = 2.
Therefore, S_{24} = (24/2)[ 2 × 5 + (24 − 1) × 2] = 12 [10 + 46] = 672
So, sum of first 24 terms of the list of numbers is 672.
 the production in the 1st year
 the production in the 10th year
 the total production in first 7 years
A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find :
 Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP. Let us denote the number of TV sets manufactured in the nth year by a_{n}.
 Now a_{10} = a + 9d = 550 + 9 × 25 = 775
So, production of TV sets in the 10th year is 775.

Also, S_{7} = (7/2)[2 × 550 + (7 − 1) × 25]]
= (7/2)[1100 + 150] = 4375
Thus, the total production of TV sets in first 7 years is 4375.
Then, a_{3} = 600 and a_{7} = 700
or, a + 2d = 600
and a + 6d = 700
Solving these equations, we get d = 25 and a = 550.
Therefore, production of TV sets in the first year is 550.
 An arithmetic progression (AP) is a list of numbers in which each term is obtained by
adding a fixed number d to the preceding term, except the first term. The fixed number d
is called the common difference.
The general form of an AP is a, a + d, a + 2d, a + 3d, . . .
 A given list of numbers a_{1}, a_{2}, a_{3}, . . . is an AP, if the differences a_{2} – a_{1}, a_{3} – a_{2}, a_{4} – a_{3}, . . ., give the same value, i.e., if a_{>k+1}k is the same for different values of k.
 In an AP with first term a and common difference d, the nth term (or the general term) is given by a_{n} = a + (n – 1) d.
 The sum of the first n terms of an AP is given by :
S = (n/2)[ 2a + (n − 1) d ]
 If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP
is given by :
S = (a + l )