The sequence of numbers like 1,2,3,4… are said to be in arithmetic progression with common difference d = 1; Generalizing this the arithmetic series is of the form: a, a+d, a+2d, a+3d… a+(n-1)d.
Common difference d is Tn – T(n-1) i.e. next term minus previous term. This is uniform throughout.
Let a denote the first term d, the common difference, and n the total number of terms. Also, let L denote the last term, and S the required sum; then
S = n(a+L)2
L = a + (n – 1)d
S = n2∗[2a+(n−1)d]
When three quantities are in arithmetic progression, the middle one is said to be the arithmetic mean of the other two.
Thus a is the arithmetic mean between a – d and a + d. So, when it is required to arbitrarily consider three numbers in AP take a – d, a and a + d as the three numbers as this reduces one unknown thereby making the solution easier.
Between two given quantities it is always possible to insert any number of terms such that the whole series thus formed shall be in AP. The terms thus inserted are called the arithmetic means
To Find the Arithmetic Mean between any Two given Quantities
Let a and b be two quantities and A be their arithmetic mean. Then since a, A, b, are in AP. We must have b – A = A – a
A = a+b2
To Insert a given Number of Arithmetic Means between Two given Quantities
Let a and b be the given quantities and n be the number of means. Including the extremes, the number of terms will then be n + 2 so that we have to find a series of n + 2 terms in AP, of which a is the first, and b is the last term.
Let d be the common difference; then b = the (n + 2)th term = a + (n + 1)d
Hence, d = (b−a)(n+1)
Thus the required means are : a+(b−a)(n+1),a+2(b−a)(n+1),a+3(b−a)(n+1),...,a+n(b−a)(n+1)
Process for finding the nth term of an A.P
Suppose you have to find the 17th term of the A.P. 3, 7, 11…….
The algorithm goes like this: In order to find the 17th term of the above sequence add the common difference to the first term, sixteen times. (Note: Sixteen, since it is one less than 17).
Similarly, in order to find the 37th term of the A.P. 3, 11 …, All you need to do is add the common difference (8 in this case), 36 times. Thus, the answer is 288 + 3 = 291.
Note : Corresponding terms of the A.P
Consider the A.P., 2, 6, 10, 14, 18, 22. If you try to find the average of these six numbers you will get: Average = (2 + 6 + 10 + 14 + 18 + 22)/6 = 12
Notice that 12 is also the average of the first and the last terms of the A.P. In fact, it is also the average of 6 and 18 (which correspond to the second and 5th terms of the A.P.). Further, 12 is also the average of the 3rd and 4th terms of the A.P
(Note: In this A.P. of six terms, the average was the same as the average of the 1st and 6th terms. It was also given by the average of the 2nd and the 5th terms, as well as that of the 3rd and 4th terms. ) We can call each of these pairs as “CORRESPONDING TERMS” in an A.P
If you try to notice the sum of the term numbers of the pair of corresponding terms given above: 1st and 6th (so that 1 + 6 = 7); 2nd and 5th(hence, 2 + 5 = 7); 3rd and 4th (hence, 3 + 4 = 7)
In each of these cases, the sum of the term numbers for the terms in a corresponding pair is one greater than the number of terms of the A.P. This rule will hold true for all A.P.s.
For example, if an A.P. has 23 terms then for instance, you can predict that the 7th term will have the 17th term as its corresponding term, or for that matter the 9th term will have the 15th term as its corresponding term. (Since 24 is one more than 23 and 7 + 17 = 9 + 15 = 24.)
Special Case : Increasing Arithmetic Progressions
Every term of an increasing AP is greater than the previous term
When the first term of an increasing A.P is negative we get a special case for some A.P’s.
Consider the following series: Series : –12, –8, –4, 0, 4, 8, 12
As is evident the sum to 2 terms and the sum to 5 terms in this case is the same. Similarly, the sum to 3 terms is the same as the sum to 4 terms. This can be written as:
S2 = S5 and S3 = S4
This doesn’t happen for all series. Series : –13, –7, –1, + 5, + 11… A clear look at the two series will reveal that this phenomenon occurs in series which have what can be called a balance about the number zero.
Series : –12, –6, 0, 6, 12 …Here we have S2 = S3 and S1 = S4. Thus where ‘0’ is part of the series) the sum is equal for two terms such that one of them is odd and the other is even.
Series : –15, –9, –3, + 3, 9, 15 …Here we have S1 = S5 and S2 = S4. Thus where ‘0’ is not a part of the series) the sum is equal for two terms such that both are odd or both are even.
Whatever was true for increasing A.Ps with first term negative will also be true for decreasing APs with first term positive.
Important Results of Arithmetic Progression :
(1 + 2 + 3 + … + n) = n * (n+1) / 2
(12 + 22 + 32 + …. + n2) = n(n+1)(2n+1)/6
(13 + 23 + 33 + …. + n3) = [n * (n+1) / 2 ] ^ 2
The series is in geometric progression if the numbers increase or decrease by a common ratio. So the series is a, ar, ar2, ar3, ar4…. arn-1
If r > 1 then Sum = a ( rn – 1) / ( r – 1)
If r < 1 then Sum = a ( 1 – rn) / ( 1 – r )
If an infinite geometric progression series is Sum = a / ( 1 – r )
When three quantities are in geometrical progression, the middle one is called the geometric mean between the other two. While arbitrarily choosing three numbers in GP, we take a/r, a and a/r. This makes it easier since we come down to two variables for the three terms.
Let a and b be the two quantities; G the geometric mean. Then since a, G, b are in GP we get G = a∗b−−−−√
To Insert a given Number of Geometric Means between two given Quantities
Let a and b be the given quantities and n the required number of means to be inserted. In all there will be n + 2 terms so that we have to find a series of n + 2 terms in GP of which a is the first and b the last. Let r be the common ratio;
Then b = the (n + 2)th term = arn+1
Three quantities a, b, c are said to be in Harmonic Progression when a/c = a−bb−c
In general, if a, b, c, d are in AP then 1/a, 1/b, 1/c and 1/d are all in HP.
There is no general formula for the sum of any number of quantities in harmonic progression. Questions in HP are generally solved by inverting the terms, and making use of the properties of the corresponding AP
Let a, b be the two quantities, H their harmonic mean; then 1/a, 1/H and 1/b are in A.P.; Then H = 2aba+b
If A, G, H are the arithmetic, geometric, and harmonic means between a and b, we have G is the geometric mean between A and H. i.e. A * H = G2
The arithmetic, geometric, and harmonic means between any two positive quantities are in descending order of magnitude. So the arithmetic mean of any two positive quantities is greater than their geometric mean and Geometric mean is greater than harmonic mean.
If the same quantity be added to, or subtracted from, all the terms of an AP, the resulting terms will form an AP, but with the same common difference as before.
If all the terms of an AP be multiplied or divided by the same quantity, the resulting terms will form an AP, but with a new common difference, which will be the multiplication/division of the old common difference.
If all the terms of a GP be multiplied or divided by the same quantity, the resulting terms will form a GP with the same common ratio as before.
If you have to assume 3 terms in AP, assume them as a – d, a, a + d or as a, a + d and a + 2d For assuming 4 terms of an AP we use: a – 3d, a – d, a + d and a + 3d For assuming 5 terms of an AP, take them as: a – 2d, a – d, a, a + d, a + 2d
To find the sum of the first n odd natural numbers = n2
To find the sum of the first n even natural numbers = n(n+1)
If we are counting in steps of x from n1 to n2 including both the end points, we get [(n2 – n1)/x] + 1 numbers.
If we are counting in steps of x from n1 to n2 including only one of the end points, we get [(n2 – n1)/x] numbers.
If we are counting in steps of x from n1 to n2 excluding both the end points, we get [(n2 – n1)/x] – 1 numbers.
Q. Count the number of terms in each case :
Between 16 and 25 both included there are (25 – 16) + 1 = 10 numbers
Between 100 and 200 both excluded there are (200 – 100) – 1 = 99 numbers
Number of numbers between 100 and 200 divisible by three ? The first number is 102 and the last number is 198. Hence, answer = ([198-102]/3) + 1 = 33
How many terms are there in the series 107, 114, 121, 128 … 254 ? (254 – 107)/7 + 1 = 147/7 + 1 = 21 + 1 = 22 terms in the series
To find how many terms of the series 107, 114, 121, 128 … are below 258, then we have by the formula: (258 – 107)/7 + 1 = 151/7 + 1 = 21.57 + 1. = 22.57. This will be adjusted by taking the lower integral value = 22.
How many numbers between 11 and 90 are divisible by 7 ?
The required numbers are 14, 21, 28, 35 ….. 77, 84, This is an A.P. with a = 14 and d = (21 – 14) = 7. If it contains ‘n’ terms then Tn = = 84 => a + (n – 1) d = 84. => 14 + (n – 1) x 7 = 84 or n = 11. Required number of terms = 11.
Find the sum of all odd numbers upto 100.
The given numbers are 1, 3, 5, 7, …, 99. This is an A.P. with a = 1 and d = 2. Let it contain n terms. Then, 1 + (n – 1) x 2 = 99 or n = 50. Required sum = n/2 * (first term + last term) = 50/2*(1+99) = 2500
Find the sum of all 2 digit numbers divisible by 3.
All 2 digit numbers divisible by 3 are : 12, 51, 18, 21, …, 99. This is an A.P. with a = 12 and d = 3. Let it contain n terms. Then, 12 + (n – 1) x 3 = 99 or n = 30. Required sum = 30/2 x (12+99) = 1665.
How many terms are there in 2,4,8,16 … 1024?
Clearly 2,4,8,16…1024 form a GP. With a = 2 and r = 4/2 = 2. Let the number of terms be n . Then 2 * 2n – 1 = 1024. So 2n = 1024 and n = 10
Two persons—Ramu Dhobi and Kalu Mochi have joined Donkey-work Associates. Ramu Dhobi and Kalu Mochi started with an initial salary of ` 500 and ` 640 respectively with annual increments of ` 25 and ` 20 each respectively. In which year will Ramu Dhobi start earning more salary than Kalu Mochi?
The current difference between the salaries of the two is ` 140. The annual rate of reduction of
this dfference is ` 5 per year. At this rate, it will take Ramu Dhobi 28 years to equalise his salary with
Kalu Dhobi’s salary.
Thus, in the 29th year he will earn more.
This problem should be solved while reading and the thought process should be 140/5 = 28. Hence,
answer is 29th year.
Find the value of the expression 1 – 6 + 2 – 7 + 3 – 8 +……… to 100 terms
we can do this faster by considering (1 – 6), (2 – 7), and so on as one unit or one term. 1 – 6 = 2 – 7 = … = –5. Thus the above series is equivalent to a series of fifty –5’s added to each other.
So, (1 – 6) + (2 – 7) + (3 – 8) + … 50 terms = –5 × 50 = –250
Find the sum of all numbers divisible by 6 in between 100 to 400
Here 1st term = a = 102 (which is the 1st term greater than 100 that is divisible by 6.)
The last term less than 400, which is divisible by 6 is 396.
The number of terms in the AP; 102, 108, 114…396 is given by [(396 – 102)/6] +1= 50 numbers.
Common difference = d = 6
So, S = 25 (204 + 294) = 12450
Find t10 and S10 for the following series: 1, 8, 15,…
This is an AP with first term 1 and common difference 7.
t10 = a + (n – 1) d = 1 + 9 × 7 = 64. S10 = 10 * (64+1)/2 = 325
Find t18 and S18 for the following series: 2, 8, 32, …
2 * 4 18
2 * 4 19
2 * 4 16
2 * 4 17
Ans .2 * 4 17
This is a GP with first term 2 and common ratio 4
Is the series 1, 4,… to n terms an AP, or a GP, or an HP, or a series which cannot be determined?
Ans .cant say
To determine any progression, we should have at least three terms.
Find the sum to 200 terms of the series 1 + 4 + 6 + 5 + 11 + 6 +…
None of these
we can treat every two consecutive terms as one.
So we will have a total of 100 terms of the nature:
(1 + 4) + (6 + 5) + (11 + 6) … = 5, 11, 17…
Now, a = 5, d = 6 and n = 100
Hence the sum of the given series is S = 100/2 * [2 * 5 + 99 * 6] = 30200
How many terms of the series –12, –9, –6,… must be taken that the sum may be 54?
-3 / 12
Here S = 54, a = –12, d = 3, n is unknown and has to be calculated. To do so we use the formula for the sum of an AP and get 54 = [2(−12)+3(n−1)]∗n2 Solving this we get n = -3 / 12. we reject negative value for n.
Find the sum of n terms of the series 1.2.4 + 2.3.5 + 3.4.6 + …
n(n + 1)(n + 2)
(n(n + 1)/12)(3n2 + 19n + 26)
((n + 1)(n + 2)(n + 3))/4
(n2(n + 1)(n + 2)(n + 3))/3
Ans .(n(n + 1)/12)(3n2 + 19n + 26)
If we put n = 1, we should get the sum as 1.2.4 = 8. By substituting n = 1 in each of the four options we
will get the following values for the sum to 1 term:
Option (a) gives a value of: 6
Option (b) gives a value of: 8
Option (c) gives a value of: 6
Option (d) gives a value of: 8
From this check we can reject the options (a) and (c).
Now put n = 2. You can see that up to 2 terms, the expression is 1.2.4 + 2.3.5 = 38
The correct option should also give 38 if we put n = 2 in the expression. Since, (a) and (c) have already
been rejected, we only need to check for options (b) and (d).
Option (b) gives a value of 38
Option (d) gives a value of 80.
Hence, we can reject option (d) and get (b) as the answer.
How many terms are there in the AP 20, 25, 30,… 130
In order to count the number of terms in the AP, use the short cut:
[(last term – first term)/ common difference] + 1. In this case it would become:
[(130 – 20)/5] +1 = 23. Option (b) is correct
Bobby was appointed to ABC in the pay scale of ` 7000–500–12,500. Find how many years he will take to reach the maximum of the scale.
7000 – 500 – 12500 means that the starting scale is 7000 and there is an increment of 500 every year. Since, the total increment required to reach the top of his scale is 5500, the number of years
required would be 5500/500 = 11. Option (a) is correct
Find the 1st term of an AP whose 8th and 12th terms are respectively 39 and 59.
Since the 8th and the 12th terms of the AP are given as 39 and 59 respectively, the difference
between the two terms would equal 4 times the common difference. Thus we get 4d = 59 – 39 =
20. This gives us d = 5. Also, the 8th term in the AP is represented by a + 7d, we get:
a + 7d = 39 = a + 7 × 5 = 39 Æ a = 4. Option (c) is correct.
A number of squares are described whose perimetres are in GP. Then their sides will be in
If we take the sum of the sides we get the perimeters of the squares. Thus, if the side of the respective squares are a1, a2, a3, a4… their perimeters would be 4a1, 4a2, 4a3, 4a4
. Since the perimeters are in GP, the sides would also be in GP
There is an AP 1, 3, 5…. Which term of this AP is 55?
The number of terms in a series are found by: [(Last term – first term)/Common difference ] + 1
How many terms are identical in the two APs 1, 3, 5,… up to 120 terms and 3, 6, 9,… up to 80 terms?
The first common term is 3, the next will be 9 (Notice that the second common term is exactly 6
away from the first common term. 6 is also the LCM of 2 and 3 which are the respective common
differences of the two series.)
Thus, the common terms will be given by the A.P 3, 9, 15 ….., last term. To find the answer you
need to find the last term that will be common to the two series.
The first series is 3, 5, 7 … 239
While the second series is 3, 6, 9 ….. 240.
Hence, the last common term is 237. Number of terms = (237 – 3)/6 + 1 = 40
Find the lowest number in an AP such that the sum of all the terms is 105 and greatest term is 6 times the least.
Trying Option (a),
We get least term 5 and largest term 30 (since the largest term is 6 times the least term).
The average of the A.P becomes (5 + 30)/2 = 17.5
Thus, 17.5 × n = 105 gives us:
to get a total of 105 we need n = 6 i.e. 6 terms in this A.P. That means the A.P. should look like:
5, _ , _ , _, _, 30.
It can be easily seen that the common difference should be 5. The A.P, 5, 10, 15, 20, 25, 30 fits the
The same process used for option (b) gives us the A.P. 10, 35, 60. (10 + 35 + 60 = 105) and in the
third option 15, 90 (15 + 90 = 105).
Hence, all the three options are correct
Find the 15th term of the sequence 20, 15, 10, …
The first term is 20 and the common difference is –5, thus the 15th term is:
20 + 14 × (–5) = –50. Option (c) is correct
A number 15 is divided into three parts which are in AP and the sum of their squares is 83. Find the smallest number.
The three parts are 3, 5 and 7 since 32 + 52 + 72 = 83. Since, we want the smallest number, the
answer would be 3.
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is
a = 5, a + 2d = 15 means d = 5. The 16th term would be a + 15d = 5 + 75 = 80. The sum of the series would be given by: [16/2] × [5 + 80] = 16 × 42.5 = 680