One of the simpler type of questions to appear in entrance tests, progressions are strictly formula based and rely on your ability to see patterns in questions. In this article, I will be covering the basics of arithmetic progression with the probable question types.
As is commonly understood, arithmetic progression is simply a series with a constant difference between any two consecutive terms. So, a series of terms 1, 2, 3, 4, 5 … n is an arithmetic progression with a common difference 1. The terminology for a generic arithmetic progression is as follows
The questions asked involve these concepts:
1) Value or notation of the nth term of the series
2) Sum of the first n terms of the series
3) Number of terms in the series
As you can see in the above progression the nth term of an arithmetic progression is given by
This is easy to remember if you understand the logic involved
Number of terms in a series is given by
Also, it would be a good idea to understand how to count numbers in a series using simple methods. For example, if you have to find the number of natural numbers between 100 and 450 that are divisible by 5 but not by 7, the logic would be to find the number of integers divisible by 5 that lie between 100 and 450, and subtract from it the number of integers that are divisible by 35.
To find the number of 5s, you can see that the smallest multiple of 5 greater than 100 is 105 and the largest multiple of 5 less than 450 is 445. Now, this is simply an arithmetic progression with T1 = 105, Tn = 445 and common difference = 5. So, number of terms is given by
Similarly you can find the number of integers divisible by 35 to be
So, the answer would be 69 – 10 = 59 terms.
Few points to remember
1) If the elements of the arithmetic progression are not ordered i.e. it is simply asked how many arithmetic progressions can be formed, then the order will not be important and (2, 4, 6) and (6, 4, 2) will be considered a single arithmetic progression
2) If the elements are ordered for eg. if there are 3 students and they get 6 chocolates such that each gets a different number of chocolates and they are in an arithmetic progression, then the number of ways will be 6 viz. (1, 2, 3) (1, 3, 2) (2, 1, 3) (2, 3, 1) (3, 1, 2) (3, 2, 1). So, be very mindful of the language of the question
3) ‘Between’ essentially means that the first and the last terms will not be considered. As we saw in the above example, 100 and 450 were not considered
4) If there are 3 terms in an arithmetic progression and you have been provided with the sum or the product of the terms, taking the terms to be a – d, a and a + d would be beneficial while calculating
5) If there are 4 terms in an arithmetic progression and you have been provided with the sum or the product of the terms, taking the terms to be a – 3d, a + d, a – d and a + 3d would be beneficial while calculating
6) If the sum of an arithmetic progression is asked for the terms starting from the mth term to the nth term, you will have to take the difference of two arithmetic progressions from T1 to Tm and from T1 to Tn
7) If the ratio of Tm:Tn = n:m, then the (m+n)th term will be 0
8) If in an arithmetic progression the sum of the first m terms is equal to n and if the sum of the first n terms is equal to m, the sum of (m+n) terms will be equal to –(m+n)
While the concept is very easy, the types of questions could be immense. However, the basic formulae remain the same. If you can understand the language of the question quickly and are careful while reading the question and calculating the answers these questions are must attempts.
Let’s see a few questions from the topic:
The sum of the first three terms of AP is 30 and the sum of the squares of the 1st and 2nd term is 116. Find the 7th term of the progression if its 5th term is known to be exactly divisible by 14.
Let the three terms be a – d, a, a + d.
a – d + a + a + d = 30
a = 10
(a – d)2 + a2 = 116
So, d = 6 or d = 14
If d = 6, the progression would be 4, 10, 16, 22, 28, 34, 40
If d = 14, the progression would be –4, 10, 24, 38, 52, 66, 80
We can see that the 5th term is divisible by 14 in the first case and so, the 7th term will be 40.
How many arithmetic progressions can be formed from the first 30 natural numbers such that the common difference is a factor of the last term?
This can be done by observation alone. If a 5 term series has to be formed, the common difference has to be at least 1 and at most 7. Let’s see this case wise
If common difference = 1, the series could be either of (1, 2, 3, 4, 5) (2, 3, 4, 5, 6) … (26, 27, 28, 29, 30) so, 26 series (remember that you can simply observe the first or the last term of each series and find out the number of series easily)
If common difference = 2, the series would have to end with an even number because of the condition given. So, the series could be either of (2, 4, 6, 8, 10) (4, 6, 8, 10, 12) … (22, 24, 26, 28, 30) so, 11 series
If common difference = 3, there would be 6 series
If common difference = 4, there would be 3 series
If common difference = 5, there would be 2 series
If common difference = 6, there would be 1 series
There cannot be any series formed with common difference = 7 as the smallest such series that satisfies the criterion would be (7, 14, 21, 28, 35) which would lie outside the first 30 natural numbers.
So, total number of series is 26 + 11 + 6 + 3 + 2 + 1 = 49
There are 30 Buzz Lightyear toys to be distributed among 3 kids. The number of toys distributed to each kid have to be in an arithmetic progression such that no kid receives more than 13 toys and each kid receives a different number of toys. In how many ways can this be done?
(a) 3 (b) 9 (c) 18 (d) 27
Let the arithmetic progression be a – d, a, a + d
So, we get
a – d + a + a + d = 30
a = 10
So, one of the kids gets 10 toys. Now, the other kids have to get 20 toys in total such that each kid gets an unequal number of toys. So, the various combinations that would satisfy the criteria of 30 toys being distributed such that each kid gets an unequal number of toys would be:
(1, 10, 19) (2, 10, 18) (3, 10, 17) … (9, 10, 11)
But, there is an additional criterion that no kid should receive more than 13 toys and so, we are left with the following cases:
(7, 10, 13) (8, 10, 12) (9, 10, 11) i.e. 3 combinations
But 3 is not the final answer because we are distributing the objects among 3 kids and so, we do not know which one get which number. Also, it would be wrong to assume that the kids receive the toys in an arithmetic progression ‘in a particular order’. So, the final answer will be nothing but 3*3! = 18 combinations.
There is an arithmetic progression consisting of terms a1, a2, a3… and another arithmetic progression consisting of terms b1, b2, b3… such that a1b1, a2b2, a3b3… is 192, 360, 576…. What will be the value of a8b8?
Let the first arithmetic progression be of the form
a, a + d, a + 2d…
and the second arithmetic progression be of the form
b, b + f, b + 2f…
ab = 192
(a + d)(b + f) = 360 => ab + af + bd + df = 360 => af + bd + df = 168
(a + 2d)(b + 2f) = 576 => 2af + 2bd + 4df = 384
So, 2df = 48, df = 24
af + bd = 144
a8b8 = (a + 7d)(b + 7f) = 192 + 7*144 + 49*24 = 2376.
A couple of CAT questions for practice:
The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e… is?
(a) u (b) v (c) w (d) x
Sum of the first 11 terms of an arithmetic progression equals the sum of the first 19 terms of that progression. What is the sum of the first 30 terms?
(a) 0 (b) -1 (c) 1 (d) Cannot be uniquely determined
Hope that clears a few doubts regarding arithmetic progressions. While knowing all the types of questions is practically impossible, one can be aware of the various traps and techniques required to solve questions. If you cover the above mentioned concepts well as a part of your prep, you should be able to deal with almost all the type of questions that CAT can hurl at you.
All the best!
PS: You can read the article on geometric progression here.