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.
Properties:
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 = \( \frac{n(a + L)}{2} \)
L = a + (n - 1)d
S = \( \frac{n}{2} * [2a + (n - 1)d] \)
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 = \( \frac{a + b}{2} \)
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 = \( \frac{(b - a)}{(n + 1)} \)
Thus the required means are : \( a + \frac{(b - a)}{(n + 1)}, a + 2\frac{(b - a)}{(n + 1)}, a + 3\frac{(b - a)}{(n + 1)} , ... , a + n\frac{(b - a)}{(n + 1)} \)
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 = \( \sqrt{a * b} \)
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 = \( ar^{n + 1} \)
\( r^{(n + 1)} = \frac{b}{a} \)
\( r = \frac{b}{a}^{\frac{1}{n+1}} \)
Three quantities a, b, c are said to be in Harmonic Progression when a/c = \( \frac{a - b}{b - c} \)
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 = \( \frac{2ab}{a + b} \)
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.
Ans .
11
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.
Ans .
2500
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
Ans .
1665
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.
Ans .
10
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
Ans .
29
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.
Ans .
-250
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
Ans .
12450
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
Ans .
325
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
Ans .
2 * 4 17
This is a GP with first term 2 and common ratio 4
Ans .
cant say
To determine any progression, we should have at least three terms.
Ans .
30200
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
Ans .
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 = \( \frac{[2(-12) + 3(n-1)]*n}{2}\) Solving this we get n = -3 / 12. we reject negative value for n.
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.
Ans .
23
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
Ans .
11
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
Ans .
4
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.
Ans .
GP
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
Ans .
28
The number of terms in a series are found by: [(Last term - first term)/Common difference ] + 1
Ans .
40
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
Ans .
all
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 situation. 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
Ans .
-50
The first term is 20 and the common difference is –5, thus the 15th term is: 20 + 14 × (–5) = –50. Option (c) is correct
Ans .
3
The three parts are 3, 5 and 7 since 32 + 52 + 72 = 83. Since, we want the smallest number, the answer would be 3.
Ans .
680
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
Ans .
a and b
If you find the sum of the series till 18 terms the value is 513. So also for 19 terms the value of the sum would be 513. Option (c) is correct.
Ans .
20
Solve this question through trial and error by using values of n from the options: For 19 terms, the series would be 5 + 8 + 11 + …. + 59 which would give us a sum for the series as 19 × 32 = 608. The next term (20th term of the series) would be 62. Thus, 608 + 62 = 670 would be the sum to 20 terms. It can thus be concluded that for 20 terms the value of the sum of the series is not less than 670
Ans .
1770
His total earnings would be 60 + 63 + 66 + … + 117 = Rs 1770. Option (a) is correct
Ans .
7
The series would be 5, 20, 80, 320, 1280, 5120, 20480. Thus, there are a total of 7 terms in the series.
Ans .
Sum of a G.P. with first term 1 and common ratio 2 and no. of terms 20. 220 - 1
Ans .
54
16r4 = 81 so r4 = 81/16 and r = 3/2. Thus, 4th term = ar3 = 16 × (3/2)3 = 54
Ans .
3
In the case of a G.P. the 7th term is derived by multiplying the fourth term thrice by the common ratio. (Note: this is very similar to what we had seen in the case of an A.P.) Since, the seventh term is derived by multiplying the fourth term by 8, the relationship. r3 = 8 must be true. Hence, r = 2. If the fifth term is 48, the series in reverse from the fifth to the first term will look like: 48, 24, 12, 6, 3. Hence, option (b) is correct
Ans .
8
Visualising the squares below 84, we can see that the only way to get the sum of 3 squares as 84 is: 22 + 42 + 82 = 4 + 16 + 64. The largest number is 8.
Ans .
7881
The series would be given by: 1, 5, 9… which essentially means that all the numbers in the series are of the form 4n + 1. Only the value in option (c) is a 4n + 1 number and is hence the correct answer
Ans .
29
The series will be 301, 308, …….. 497. So total = 196/7 + 1 = 29
Ans .
both
The answer to this question can be seen from the options. Both 2, 6, 18 and 18, 6, 2 satisfy the required conditions- viz: GP with sum of first and third terms as 20
Ans .
99
The 33rd term of the sequence would be the 17th term of the sequence 3, 9, 15, 21 ….The 17th term of the sequence would be 3 + 6 × 16 = 99
Ans .
1595
The sum to 33 terms of the sequence would be: The sum to 17 terms of the sequence 3, 9, 15, 21, …99 + The sum to 16 terms of the sequence 8, 13, 18, 83. The required sum would be 17 × 51 + 16 × 45.5 = 867 + 728 = 1595.
Ans .
637.5
The maximum sum would occur when we take the sum of all the positive terms of the series. The series 25, 24.5, 24, 23.5, 23, ….…. 1, 0.5, 0 has 51 terms. The sum of the series would be given by: n × average = 51 × 12.5 = 637.5
Ans .
12
The series would be 8, 8/3, 8/9 and so on. The sum of the infinite series would be 8/(1 – 1/3) = 8 × 3/2 = 12. Option (a) is correct
Ans .
11111
It can be seen that for the series the average of two terms is 2, for 3 terms the average is 3 and so on. Thus, the sum to 2 terms is 22, for 3 terms it is 32 and so on. For 11111 terms it would be 111112 = 123454321. Option (d) is correct