While solving problems, we generally come across terms which are in arithmetic progression, but we are not able to identify it. We can find the sum of terms, number of terms of the series if familiar with the concepts of AP.
For example, 3, 5, 7, 9, 11, 13,………is an example of an Increasing arithmetic progression with the common difference of 2. Similarly, 3, 0, –3, –6, –9,………is also an example of arithmetic progression with common difference of (–3). So, when the common difference is positive, the AP is increasing while the AP is decreasing when the common difference is negative.
We usually denote the first term by ‘a’ and if the common difference is taken as‘d’, the second term would be (a + d), third term would be (a + d + d) or ‘a + 2d’ and so on we can find the following terms.
So, first term = t1 = a
Second term = t2 = a + d
Third term = t3 = a + 2d
Fourth term = t4 = a + 3d
Fifth term = t5 = a + 4d
………………
……………..
Finally nth term = tn = a + (n – 1)d, where a is the first term, n is the number of terms and d is the common difference.
So, this is the first formula encountered in the arithmetic progression. The nth term or tn can also be understood as the last term of the arithmetic progression.
E.g. 1: If the first term of an AP is 7 and the common difference is 4, find the 25th term of this series.
This is just the direct application of the formula learnt above.
We know a = 7, d = 4, and we have to find t25.
On substituting values in the formula, we get t25 = a + 24d.
So, t25 = 7 + 24 × 4 = 7 + 96 = 103.
E.g. 2: Find the number of terms in the series: 9, 13, 17, 21, 25,………………………..609.
Again, this question is also a direct application of previous logic. Just observe which all data has been given. The first term ‘a’ = 9, common difference‘d’ = 4 and the last term ‘tn’ = 609.
Just substitute these values in the formula, we can find the value of number of terms ‘n’.
tn = a + (n – 1)d
So, (n – 1)d = tn – a
E.g. 3: Find the number of terms common to both the arithmetic progressions:
AP 1: 10, 13, 16, 19,………………………..up to 45 terms.
AP 2: 2, 6, 10, 14,…………………………… up to 40 terms.
This is a pretty easy question, but students get confused in how to start. Let us see how to solve it.
First of all, search for the first common term in both series.
First common term in both series is 10. Now, we need to determine which will be the 2nd common term present in both the series. If we are aware of the logic of LCM, we can easily guess which will be the 2nd common term.
The first AP is increasing with a common difference of +3, while the second AP is increasing with a common difference of +4. Their common meeting point will occur when the cycles of 3 and 4 will be complete. So, they will meet at 12, then at 24, then at 36 and so on.
Pay attention here: 3, 6, 9, 12, 15, 18, 21, 24, 27,….(increasing in steps of 3) and so on.
And 4, 8, 12, 16, 20, 24, 28,……..(increasing in steps of 4) and so on.
You can observe that the two series are meeting at 12, then at 24 and then they will meet at 36 and so on. This is a very handy logic which is used quite frequently.
So, coming back to our original question, we finally got the funda that two series will have common elements after every 12 or its multiples.
So, the first common term in both series was 10, then, next common term will occur after 12 means 10 + 12 = 22. Third common term will be 22 + 12 = 34.
Now, the problem is up till which limit should we find out common terms.
For this, we need to find out the last terms of both the series.
We know the first term of first AP and its common difference, so we can find out which will be the last term of the first series.
So, t45 = a + 44d = 10 + 44 × 3 = 142.
Similarly, we can also find out which will be the last term of the 2nd AP.
So, t40 = a + 39d = 2 + 39 × 4 = 158.
The advantage of finding out the last terms of both APs is that it will help us in determining up till which limit we need to find out common terms. So, we will look for common terms up till 142.
Now, how to count how many common terms are there in the series 10, 22, 34, 46, 58, ……up till 142.
This is a standard counting problem. If you do not know the logic of counting, learn it now. Its very easy.
Logic: The common difference was of 12 in the series of common terms, so divide each term of the common series by 12 and try to write that number in form of [12 × n – m] OR [12 × n + m], where n and m are integers.
Now, observing the sequence we got, there are 12 common terms present in both the series. [11 – 0 + 1 = 12 terms].
So, in total 12 terms are common in both the series.
Alternative method of counting:
First term common to both the series is 10, then 22, then 34 and we have to go up till 142. We can use the formula of AP {nth term = tn = a + (n – 1)d}.
So, number of common terms will be n= (142 – 10)/12 + 1 = 12 terms.
But, if the last term is not the common term of both the series, then first method will give us answer faster than any other method.
Sum of the first ‘n’ terms of an AP:
Now, we will learn how to find sum of the first ‘n’ terms of the arithmetic progression.
We denote sum of the first ‘n’ terms of an AP by Sn.
So, Sn = t1 + t2 + t3 + t4 + t5 + t6 + t7 + ……………………..+ tn.
This can be further broken down into Sn = (a) + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d) + (a + 6d) +……………………….+ [a + (n – 1)d].
On adding, we get ‘a’ n times so, we get: Sn = na + d [1 + 2 + 3 + 4 + 5 + 6 + ...............+ (n – 1) times].
E.g. 4: Find the sum of the first 25 terms of the series: 12, 17, 22, 27,……………………..
We know the first term as 12, common difference as 5 and number of terms as 25. Just substitute these values in the formula obtained above.
So, Sn = (25/2)[24 + 24 × 5] = 25 × 72 = 1800.
Intuitive way of looking at sum of first ‘n’ terms of AP:
Let us take a simple AP: 2, 4, 6, 8, 10, 12, 14.
Average = [first term + last term]/2 = [2 + 14]/2 = 8
Or, Average = [second term + second last term]/2 = (4 + 12)/2 = 8
Or Average = [Third term + third last term]/2 = (6 + 10)/2 = 8.
So, either ways, we can find the average and if we got the value of average, the sum of the entire series can be obtained by multiplying the average by number of terms.
So, two cases arise:
a) The numbers of terms are odd, and then the average will be the middle term of the series.
b) And if the numbers of terms are even, the average will be the sum of any two terms which are equidistant from the centre divided by 2.
So, this is a very novel way of looking at the sum of ‘n’ terms of AP. So, we will keep both the formulas in mind, but will use the second formula more frequently as it yields result faster than the previous one.
E.g. 6: Find the sum to the given series: 11, 15, 19, 23,………………….., 99, 103.
Now, in this problem, we do not know the ‘number of terms’, so we should first find out the number of terms.
We already know the faster approach of finding out the number of terms in an AP as [last term – first term]/common difference + 1.
So, ‘n’ = [103 – 11]/4 + 1 = 24 terms.
If the number of terms are ‘even’, then the average of the series is [first term + last term]/2 or sum of any two terms which are equidistant from the center divided by 2.
So, average of the series = [11 + 103]/ 2 = 57.
So, sum to first 24 terms S24 = number of terms × average = 24 × 57 = 1368.
E.g. 7: If the rth term from the beginning of an AP is 25 and rth term from the end of the same AP is 45. Then find the sum of the first 50 terms of the series.
Again, this problem is an application of the logic of average of terms. We know the value of the terms which are equidistant, so average of the AP series will be the summation of those two terms divided by 2.
So, average of the series = [25 + 45]/ 2 = 35.
And sum to first 50 terms, S50 will be = 50 × 35 = 1750.
I hope all of you are getting the logic of average funda to find the sum of first ‘n’ terms of an AP, the average funda is the crux of the entire AP. If we can assimilate that logic, we will have a crystal-clear picture of how the terms are being added in an AP and solve those questions of AP orally.
Further applications of Arithmetic Progression:
E.g. 8: If the mth term of an AP is n and nth term of the same AP is m, then find the (m + n)th term of the same AP.
Standard way:
We know, tm = a + (m – 1)d and the value of mth term is given as n.
So, a + (m – 1)d = n (Equation (i))
Similarly, we will also get a + (n -1)d = m (Equation (ii))
Now, we can solve these two equations. Subtract these two equations, we get, md – nd = n – m.
On further solving, we get d = –1 and plug-in value of‘d’ in any of the equations to find the value of ‘a’.
On substituting d = –1 in first equation, we get the value of a = n + m – 1.
Now, we need to find the value of tm+n = a + (m + n – 1)d.
Substitute the value of a = m + n – 1 and d = –1 in the equation above, we will get tm+n = 0.
Practical Way:
Assume m = 1 and n = 2, and see how easily we can solve this question.
Its given, tm = n means t1 = 2 and tn = m means t2 = 1.
So, the first term is 2 and second term is 1 which implies that common difference of the AP is (–1).
We need to find tm+n term which is t1+2 term.
So, t3 = a + 2d = 2 + 2 × (-1) = 0
So, we got the same answer and had we taken any other values of m and n, still we would have got the same answer. So, in all such problems, practical way of solving is to assume values and go ahead. Assuming values makes our job easy and calculations become easier.
E.g. 9: How many minimum terms of the AP series should be taken so that the sum of terms [–36, –32, –28,.....................................] becomes positive?
This is a very simple question which can be solved orally. In all such problems, we should be just looking for one thing that the series is increasing from negative to positive, so whether 0 will be a part of the series or no.
If yes, then the question can be solved without any pen or pencil.
If we get a 0 in the series, then whatever were the negative terms, we will also get the terms with the same magnitude but with the opposite sign.
So, when we add all these terms, we will get a sum of zero. But, the question is asking to find the sum of terms which should be positive means > 0. So, we will have to take one more term which will be 40 to make the sum of terms positive.
Now, we need to count how many terms are there. There are multiple ways of counting.
36 = 4 × 9
32 = 4 × 8
…..
…..
4 = 4 × 1.
So, there are 9 multiples of 4. But there are 9 multiples in negative part also. So, a total of 18 terms, one term will be 0 and another will be 40.
So, a total of 20 minimum terms should be taken so that sum becomes greater than 0.
Alternative way: (36 – 4)/4 + 1 = 9 terms in the positive and 9 terms in the negative. So a total of 18 terms, and two more terms which are 0 and 40. So, a total of 20 terms should be taken.
E.g. 10: If the terms of a series are 46, 43, 40, 37,……………………….. Then find how many minimum terms should be taken so that the sum of those terms becomes negative?
This is a slightly tricky question as compared to the previous one because in this question when we are decreasing from 46, 43, 40,……………….., we will not get 0 in this series. So, this will create a problem in finding out how many terms should be taken so that sum becomes less than zero.
So, the best way is to use the standard way.
So, Sn < 0 as we know the first term is 46 and common difference is (–3).
Now, this becomes a standard inequality problem which all of us should know how to solve.
So, on further solving, we get: n[3n – 95] > 0.
So, we get n > 31.66 OR n < 0.
So, the minimum number of terms that should be taken is 32. [Number of terms cannot be negative or in fractions].
So, we need to learn the approach in all such questions. If on observing the series, we find out that 0 will appear in the series, then that problem can be solved orally as all the negative terms and their positive counterparts will get cancelled when we add them.
But, if in a series, 0 does not appear, then the better way will be to apply the standard formula of Sn < 0 OR Sn > 0 depending on what the question is asking.
E.g. 11: If three roots of cubic equation x3 – 12x2 + kx – 48 = 0 are in arithmetic progression, find the value of ‘k’.
Again, this question is an application of arithmetic progression. Whenever three terms are in AP, we should assume values as (y –d), y and (y + d) which will make our job easy as when we add these three terms, ‘d’ will get canceled out and we can get the value of ‘y’.
So, let our roots of cubic equation be [y – d, y, y + d].
We know sum of roots of a cubic equation is –b/a = – (–12)/1 = 12. [Where ‘b’ is the coefficient of x2 and ‘a’ is the coefficient of x3].
So, we can equate [y – d + y + y + d] = 12. On solving we get y = 4.
Now, we know product of roots of a cubic equation is –d/a = – (–48)/1 = 48. [Where‘d’ is the constant term in the cubic equation and ‘a’ is the coefficient of x3].
So, we can equate (y – d) × (y) × (y + d) = 48.
On substituting values of ‘y’ = 4 in the equation we get {4 – d} × 4 × {4 + d} = 48.
On solving it, we get d = +2 OR – 2.
So, we finally get value of (y – d) = 2, y = 4 and (y + d) = 6.
Now, the question is asking us to find sum of roots taken two at a time = c/a [where ‘c’ is the coefficient of x].
So, (y – d) × (y) + (y) × (y + d) + (y – d) × (y + d) = k
Just plug values of y = 4, y – d = 2 and y + d = 6.
We get, [2 × 4 + 4 × 6 + 2 × 6] = k
Finally, k = 44.
E.g. 12: If the sum of the first 20 terms of an AP is equal to the sum of the next 15 terms of the same series. If the 44th term of the same series is 172, find the common difference of the AP.
Again a standard problem based on the funda of sum of first ‘n’ terms but with some twist in it.
And be careful of the sentence “sum of the next 15 terms” is not same as sum of first 15 terms. We know how to find the sum of the first 15 terms, but how do we find sum of next 15 terms.
E.g. 14: Find the sum of all the two-digit numbers which when divided by 5 leave remainder of 2.
The general form of all such numbers which when divided by 5 leave remainder of 2 is (5n + 2).
So, first two-digit number which will satisfy this condition 12 then 17, 22, 27 and up till 97. This forms an AP with the common difference of 5 and first term as 12.
We need to find the number of terms to get the desired result.
Number of terms = (97 – 12)/5 + 1 = 18 terms.
Now Sum of these 18 terms = Number of terms × Average.
Sum = 18 × [t1 + t18]/2.
Sum = 18 × [12 + 97]/2
On solving, we get sum = 9 × 109 = 981.
Exercise:
1. Find the sum of the first 33 terms of the series: 5, 12, 19, 26,…………………
2. If 7 times the 7th term of an AP is equal to the 11 times the 11th term of same AP. Find the 18th term of the series.
3. How many minimum terms of an AP [–56, –53, –50, –47,..................] should be taken so that the sum of the terms becomes positive?
4. Find the value of ‘P’ so that P + 3, 3P – 6 and 2P – 3 are in AP.
5. If the pth term of an AP is 1/q and the qth term of the same AP is 1/p, then find the value of (pq)th term of the same AP.
6. The sum of the 5th and 18th terms of an AP is equal to the sum of the 7th, 12th and 17th terms of the same series. Then which element of the series should necessarily be equal to zero?
7. Let X be the set of integers {11, 19, 27, 35,…………………..451, 459} and Y is a subset of X such that sum of no two elements of Y is 470. Find the maximum number of elements in set Y.
8. What is the sum of the all two-digit numbers that give a remainder of 3 when they are divided by 7?
9. If three positive real numbers p, q and r satisfy q – p = r – q and pqr = 4, then find the minimum possible value of q?
10. If the pth term of an AP is 4p + 1, then find the common difference of the AP.
11. The sum of the 12 terms of an AP, whose first term is 8, is 240. Find the last term of the series?
12. Find the sum of 15 terms of an AP whose middle term is 40.
13. Find the sum of the following series: [1 – 4 + 2 – 5 + 3 – 6 + .......................to 150 terms].
14. The sum of an arithmetic progression consisting of 12 terms, is 354. The ratio of the sum of the odd terms to the sum of the even terms is 27 : 32, find the common difference of the series?
15. Find the sum of the first 13 terms of an AP whose nth term is (5n + 1).
16. The number of all the terms of an AP is odd and if the sum of all the terms is 75 and number of terms is 10 more than the middle term, find the middle term.
17. If the sum of the first 2n terms of the AP 2, 5, 8, 11,……. is equal to the sum of the first n terms of the AP 57, 59, 61,…………., then find the value of n?
18. Find the 280th term of the series: a, b, b, c, c, c, d, d, d, d, e, e, e, e, e,……………………
