Sequence & Series Concepts & Solved Examples for CAT  Nitin Gupta, AlphaNumeric

A sequence is a set of numbers written in a particular order. We sometimes write u1 for the first term of the sequence, u2 for the second term, and so on. We write the nth term as un.
A series is a sum of the terms in a sequence. If there are n terms in the sequence and we evaluate the sum then we often write Sn for the result, so that Sn = u1 + u2 + u3 + . . . + un .
An arithmetic progression, or AP, is a sequence where each new term after the first is obtained by adding a constant d, called the common difference, to the preceding term. If the first term of the sequence is a then the arithmetic progression is a, a + d, a + 2d, a + 3d, . . . where the nth term is a + (n − 1)d.
sum of n terms of an AP = n/2 ( 1st term + last term ) = n/2 [2a+(n1)d]if 25th term of an AP is 50, then find the sum of first 49 terms of the series
Always remember  if no. of terms in a series are odd; then middle term will always represent average ; so here 25th term is the middle term so sum is 49 * 50 = 2450 ( you can do it mentally & can save some time)
If the sum of first 12 numbers is equal to sum of 1st 18 numbers, then find the sum of first 30 terms of this AP.
by formula
(a + 11d) * 6 = (a + 17d) * 9
=> 3(a + 29d) = 0
Hence, sum till 30th term = 0Another important concept is that here if sum of 12 = sum of 18 > sum of 13,14,15,16,17 & 18th term is zero > 13,14,1516,17,18 > middle lines represent the middle point > sum of 30 terms is zero.
The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?
(1) 1st
(2) 9th
(3) 12th
(4) None of the abovea + 2d + a + 14d = a + 5d + a + 10d + a + 12d
=> a + 11d = 0 > 12th termIf 9 times 9th term of an AP = 13 times 13th term of same AP. which term is Zero?
9(a + 8d) = 13(a + 12d)
4a + 84d = 0 => 22 term !The third term of a finite series in Arithmetic Progression is 28. The sum of the first three terms is 54. The first term of the series is:
(1) 8
(2) 10
(3) 18
(4) 2Let the first term of an A.P. be ‘a’ and the common difference be ‘d’.
Then the second term would be a + d and third term would be a + 2d.
∴ a + (a + d) + (a + 2d) = 54
∴ 3a + 3d = 54
∴ a + d = 18 … (i)
And a + 2d = 28 … (ii)
Solving (i) and (ii) for a and d, we get a = 8 and d = 10
∴ The first term of this A.P. would be 8.How many terms are identical in the arithmetic progressions 36, 72, 108, 144, … , 1584 and 48, 96, 144, 192, … , 1680?
The first number which is common in both the A.P. is 144 and the number 144 is fourth number in the first progression. in the first series there are 44 terms; so no. of terms = 44/4 = 11
The number of common terms in the two sequences 17, 21, 25, … , 417 and 16, 21, 26, … , 466 is [CAT 2008]
(1) 78
(2) 19
(3) 20
(4) 77
(5) 22LCM 20 (see the common difference of both APs....one has difference 4 nd other has 5 so LCM OF 4,5 = 20)
Insert five arithmetic means between 8 and 26
Let A1, A2, A3, A4 and A5 be five arithmetic means between 8 and 26.
Therefore, 8, A1, A2, A3, A4, A5, 26 are in A. P. with a = 8, b = 26, n = 7
We have 26 = 8 + (7 – 1) d
d = 3
A1 = a + d = 8 + 3 = 11
A2 = a + 2d = 8 + 2 × 3 = 14
A3 = a + 3d = 17
A4 = a + 4d = 20
A5 = a + 5d = 23
Hence, the five arithmetic means between 8 and 26 are 11, 14, 17, 20 and 23.The n AM's between 20 and 80 are such that the ratio of the first mean and the last mean is 1 : 3. Find the value of n.
let means be x and 3x...x + 3x = 20 + 80=100, so x =25....so numbers are 20 , 25, ....75, 80..hence 11.
Given the set of n numbers, n > 1, of which one is 1 − (1/n), and all the others are 1. The arithmetic mean of the n numbers is [FMS 2010]
(1) 1
(2) n  1/n
(3) n  1/n^2
(4) 1  1/n^2**The angles of a pentagon are in arithmetic progression. One of the angles, in degrees, must be [FMS 2010]
(1) 108
(2) 90
(3) 72
(4) 54sum of angles of pentagon = 540/5
sum of interior angles of polygon = 180(n2)Because of economic slowdown, a multinational company curtailed some of the allowances of its employees. Rashid, the marketing manager of the company whose monthly salary has been reduced to Rs.42000 is unable to cut down his expenditure. He finds that there is a deficit of Rs.2000 between his earnings and expenses in the first month. This deficit, because of inflationary pressure, will keep on increasing by Rs.500 every month. Rashid has a saving of Rs.60000 which will be used to fill this deficit. After his savings get exhausted, Rashid would start borrowing from his friends. How soon will he start borrowing? [IIFT 2009]
(1) 10th month
(2) 11th month
(3) 12th month
(4) 13th month60000 = n/2[2*2000 + (n1) * 500]
n^2 + 7n  240 = 0
now solve for n and get the answer
n will be > 12 ...so 13 is the answerA group of 630 children is arranged in rows for a group photograph session.Each row contains three fewer children than the row in front of it. What number of rows is not possible?
(1) 3
(2) 4
(3) 5
(4) 6
(5) 7Let there be n rows and a students in the first row.
∴ Number of students in the second row = a + 3
∴ Number of students in the third row = a + 6 and so on.
∴ The number of students in each row forms an arithmetic progression with
common difference = 3
The total number of students = The sum of all terms in the arithmetic progression ;
now check with options....its 6.Consider the set S = {1, 2, 3, …, 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements? [CAT 2008]
(1) 3
(2) 4
(3) 6
(4) 7
(5) 8Let there be n terms (n ≥ 3) in the arithmetic progression having 1 as the first term and 1000 as the last. Let d be the common difference. Then, 1000 = 1 + (n – 1) × d
∴ 999 = (n – 1) × d ... (i)
∴ Factors of 999 are 1, 3, 9, 27, 37, 111, 333 and 999
Substituting in equation (i)
If d = 1, n = 1000
If d = 3, n = 334
If d = 9, n = 112
If d = 27, n = 38
If d = 37, n = 28
If d = 111, n = 10
If d = 333, n = 4
If d = 999, n = 2, which is not possible as n > 2
∴ 7 arithmetic progressions can be formed.
Hence, option 4.if a1, a2, a3, ..... is an A.P. such that a1 + a5 + a10 + a15 + a20 + a24 = 225. then a1+ a2 + a3 + .... + a23 + a24 is equal to
a. 902
b. 75
c. 750
d. 900a1 + a24 = a5 + a20 = a10 + a15 (let this be x) , so 3x = 225 implies x = 75 and we need to find 12x which is 12 x 75 = 900
OR
there are 24 term so a1+a24 = a2+a23=a3+22=....; & sum of 3 pairs = 225 ;& there r 12 pairs so sum is (225/3) *4 = 900
Consider an A.P. with 1st term 'a' and common difference 'd'. let sk denote the sum of first k terms. if S(kx)/S(x) is independent of x, then
(1) a =2d
(2) a= d
(3) 2a=d
(4) none of these.s(kx) = (kx/2)[2a+(kx1)d] = (kx/2)(2ad+kxd]
s(x) = x/2)(2a d + x * d )
so 2a = d.The interior angle of polygon are in AP. the smallest angle is 120 & the common difference is 5. find the number of sides of the polygon.
Exterior angles are also in Ap :a reducing ap start from 60 decrease by 5 so 60+55 + .....+20(9 terms ) will give us 360 degrees ...sum of exterior angles.....so 9 sides
The sum of n terms terms of two series in AP are in the ratio (3n13):(5n+21). find the ratio of their 24term.
n/2[2a+(n1)d1]/n/2[2a2+(n1)d2] = 3n13/5n+21> 2a1+(n1)d1/2a2+(n1)d2 = 3n13/5n+21;
now 24 term of first ap/24th term of 2nd ap = a1+(241)d1/(a2+(241)d2 ;
on putting n1 46 u will get the answerPlease go through this 10 mins video and then we will solve more problems from GP.
Which term of the GP : 2, 8, 32, ... up to n terms is 131072?
Let 131072 be the nth term of the given G.P. Here a = 2 and r = 4.
Therefore 131072 = an = 2(4)^(n – 1) or 65536 = 4^(n – 1)
This gives 48 = 4^(n – 1)
So that n – 1 = 8, i.e., n = 9. Hence, 131072 is the 9th term of the G.P.** How many terms in the geometric progression
1, 1.1, 1.21, 1.331, . . . will be needed so that the sum of the ﬁrst n terms is greater than 20?**The sequence is a geometric progression with a = 1 and r = 1.1. We want to ﬁnd the smallest value of n such that Sn > 20. Now by applying the formulae u will get 1.1^n > 3 ; now here basically you need to find how many successive change of 10% will be more than 200%; u will get n = 12
How many terms in the GP 4, 3.6, 3.24, . . . are needed so that the sum exceeds 35?
here a = 4, r = 9/10 so sum 4(1(9/10)^n)/(19/10) > 35 > 1/8 >(9/10^n) >.125> .9^n ; so here there is basically decrease of 10% & 20 successive change will comes out to be less than .125; so ans is 20
Find the sum of the sequence 7, 77, 777, 7777, ... to n terms
7 + 77 + 777 + 7777 + ... to n terms
=7/9 [9 + 99 + 999 + 9999 to n terms]
= 7/9[(10 1)+ (10^2 1) (10^3 1) (10^4 1) ... n terms]; solve this u will get the answer.Find the sum of n terms of the series 5 + 55 + 555 + 5555 + ...
Sn = 5 + 55 + 555 + 5555 + ... + n terms
= 5 (1 + 11 + 111 + 1111 + ... )
= 5/9[(101)+(1001)+(10001) + ... n terms]
now just separate n solveTwo consecutive numbers from 1,2,3......n are removed arithmetic mean of the remaining numbers is 105/4..Find n those removed numbers
Let p and p+1 be removed number from 1,2,.......,n then sum of the remaining numbers n(n+1)/2  (2p+1)
from given condition 105/4 = {n(n+1)/2  (2p+1)}/n2
=>2n^2103n 8p+206 = 0
since n and p are integers so n must be even let n = 2r
so we get p = {4r^2+103(1r)}/4
Since p is an integer then 1r must be divisible by 4..Let r = 1+4t ,we get
n = 2+8t and p = 16t^2  95t +1, 1 1 t = 6
=>n = 50 and p = 7
so removed numbers are 7 and 8Find out the largest term of the sequence 1/503 , 4/524 , 9/581 ,16/692,.....
You can say general term is of form (n^2)/500+3n^3
let Un = 1/tn = 500/n^2 +3n
then dTn/dn = {n(1000  3n^3)}/(500+3n^3)^2
for min n max of Tn dTn/dn = 0
so n = (1000/3)^1/3
now 6 < (1000/3)^1/3 < 7
so largest term of the sequence is 49/1529
hope its fully clear nowInsert three numbers between 1 and 256 so that the resulting sequence is a G.P.
Let G1, G2,G3 be three numbers between 1 and 256 such that
1, G1,G2,G3 ,256 is a G.P.
Therefore 256 = r^4 giving r = ± 4 (Taking real roots only)
For r = 4, we have G1 = ar = 4, G2 = ar2 = 16, G3 = ar3 = 64
Similarly, for r = – 4, numbers are – 4,16 and – 64.
Hence, we can insert, 4, 16, 64 or – 4, 16, –64, between 1 and 256 so that the resulting sequences are in G.P.From a set of first 20 natural no. 3 numbers are selected such that they are in gp. in how many ways it can be done.
1,2,4;(5 multiples of this series) ;1,3,9(2 multiples); 1,4,16 (1 ways); now when r = fraction ,now when r = fraction first term must be perfect square or it's multiple; e.g.p^2/q^2,p/q,1 i.e for r = 3/2> 4,6,9(2multiples) ;r =5/2 >4,10,25(not possible); r=4/3> 9,12,15, r= 5/3>9,15,25(not possible); r= 5/4 > not possible...so total is 11*2=22
Now watch below video for AGP fundamentals
Find the sum of the following arithmetico geometric sequence: 2, 5x, 8x^2, 11x^3, ... upto 'n' terms
The given arithmetico geometric sequence is formed from the arithmetic sequence 2, 5, 8, 11, .. and the geometric sequence 1, x, x^2, x^3, ...
Therefore for the above arithmetico geometric sequence,
First term of arithmetic sequence, a = 2
Common difference of arithmetic sequence, d = 3
Common ratio of geometric sequence, r = x
nth term of the above arithmetico geometric sequence is: Tn = (a + (n  1)d)r^(n  1)
Substituting the values of a, d and r, we get: Tn = (2 + (n  1)3)x^(n  1)
Simplifying, we get Tn = (3n  1)x^(n  1)
and (n  1)th term of the above sequence is Tn  1 = (a + (n  2)d)r^(n – 2)
That is, Tn  1 = (2 + (n  2)3)x^(n – 2) = (3n  4)x^(n – 2)
Now we can write the sum of the artihmetico geometric sequence as
Sn = 2 + 5x + 8x2 + 11x3... + (3n  1)x^(n  1)
Multiplying the above sum by common ratio 'r',
x Sn = 2x + 5x^2 + 8x^3 + ... + (3n  4)x^(n – 2) + (3n  1)xn
Subtracting the above two equations,
Sn(1  x) = 2 + 3x + 3x^2 + 3x^3 + ... + (3n  4)x^(n – 1)  (3n  1)xn
In the above equation, the terms ranging from 3x to (3n  4)x^(n – 1) form a geometric series, therefore we can write the above equation as,
2/(1x)+3x(1x^(n1))/(1x)^2((3n1)x^n/(1x)Find the sum to infinite terms: y + 4y^2 + 7y^3 + 10y^4 + 13y^5 + ...
The series given above is an AGP. We can easily identify an AGP as one term of the series will be in AP
while the other one will be in GP.
In this series y, y^2, y^3, y^4, y^5 are in GP with a common ratio of ‘y’, while the terms 1, 4, 7, 10, 13 and so
on are in AP.
So, the standard way of solving AGP problems is to multiply the entire series by the common ratio of the
GP and then subtract the newly formed series from the original series. Why do we subtract?
Because on subtracting, we will get a pure geometric progression of infinite terms. And, we know how
to find the sum of infinite terms of a GP.
So, let P = y + 4y^2 + 7y^3 + 10y^4 + 13y^5 + ........................[equation (i)]
Now, the common ratio of the GP in AGP is ‘y’. So, multiply the series by ‘y’ .
On multiplying, we get Py = y^2 + 4y^3 + 7y^4 + 10y^5 + 13y^6 +.......................[equation (ii)]
Now, the trick while subtracting is that leave the first term of the equation (i) as it, start subtracting
from the second term onwards as they are like terms.
So, on subtracting, we get:
P – Py = y + (4y^2 – y^2) + (7y^3 – 4y^3) + (10y^4 – 7y^4) +......................
P(1 – y) = y + 3y^2 + 3y^3 + 3y^4 + ........................
Now, starting from the 2nd term onwards, the series becomes an infinite GP with first term being 3y^2
and the common ratio being ‘y’.
So, we can find the sum of infinite terms using the formula [a/(1 – r)].
So, on further solving we get,
y+2y^2/(1y)^2Find the sum of series 1 * 2 + 2 * 2^2 + 3 * 2^3 +…+ 100 * 2^100
Let S = 1 * 2 + 2 * 2^2 + 3 * 2^3 +…+ 100 * 2^100 …… (1)
⇒ 2S = 1 * 2^2 + 2 * 2^3 +…+ 99 * 2^100 + 100 * 2^101 …… (2)
⇒ –S = 1 * 2 + 1 * 2^2 + 1 * 2^3 +…+ 1 * 2^100 – 100 * 2^101
⇒ –S = 1 * 2 (2^(100–1)/2–1) – 100 * 2^101
⇒ S = –2^101 + 2 + 100 * 2^101 = 99 * 2^101 + 2The infinite sum 1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^5 .....
we can take 1/7=x then try...and finally substitute x=1/7 S=1/(1x)^2+2x/(1x)^3....we get 392/216=49/27
Find the nth term of the series 1 + 2 + 5 + 12 + 25 + 46 + ....
We will solve this by another approach(coefficient method)
1 2 5 12 25 46 ...........
1 3 7 13 21 (1st consequtive diff)
2 4 6 8 (2nd consequtive difference)
2 2 2 (constant)
then Tn = a(n1)(n2)(n3)+b(n1)(n2)+c(n1) +d
putting n = 1,2,3,4 we get
T1 = d = 1,T2= c+d = 2, T3= 2b+2c+d = 5
and T4 = 6a+6b+3c+d = 12
so a= 1/3, b = 1, c =1, d = 1
Hence Tn = (n/3)*(n^2  3n +5)Find the nth term of the series 3 + 7x + 13x^2 + 21x^3 + ....
Again, we will use coefficient method
the difference of a term n its previous term as follows:
4 , 6 , 8 , ...
2 2 ...
here we see that difference is same after two operations
hence Nth term is of form Un = an^2+bn+c
where a,b,c are constants ,for n = 1,2,3
3 = a+b+c  (1)
7= 4a+2b+c (2)
13 = 9a+3b+c(3)
solving the 3 equations we get a=1,b=1,c=1
so Un = (n^2+n+1)
hence n th term of the given series is (n^2+n+1) * x^n1IF 3 + 5R + 7R^2 +..............INFINITY = 44/9 , FIND "R"
Sn = 3 / 1R + 2R/(1R)^2 = 44/9
=> (3R)/(1R)^2 = 44/9
Now denominator should be 3^2...
So by trial and error got (R=1/4)if 3 + (3+d).1/4 +(3+2d).1/16 +.........infinity = 8 , find d
Sn = (a/1r)+dr/(1r)^2
now just plugin the values n solve..check
(3/11/4) + d*1/4/(11/4)^2
=> 4 + 4d/9 = 8
=> d = 9