Lets start with Progressions...!!! I will post Questions from this topic which made their presence in Mocks and previous CAT papers....!!!! 😃
Progressions SET-1 :-
1) The sum of three numbers in A.P. is 45. If the sum of their squares is 683, what is the largest of the three numbers?
(a) 16 (b) 19 (c) 17 (d) 18
2) If (7)^1/p = (49)^1/q = (2401)^1/r , then p, q and r are in
(a) A.P. (b) G.P. (c) H.P. (d) Cannot be determined
3) The letters of the English alphabet, in the order A to Z, are made to represent 26 numbers which are in Arithmetic Progression. The sum of the numbers representing A, C and E is 36 while that of A, C, E and G is 60. What is the sum of the numbers representing B, D, F and H?
(a) 96 (b) 66 (c) 72 (d) 84
Progressions SET-1 :-
1) The sum of three numbers in A.P. is 45. If the sum of their squares is 683, what is the largest of the three numbers?
(a) 16 (b) 19 (c) 17 (d) 18
2) If (7)^1/p = (49)^1/q = (2401)^1/r , then p, q and r are in
(a) A.P. (b) G.P. (c) H.P. (d) Cannot be determined
3) The letters of the English alphabet, in the order A to Z, are made to represent 26 numbers which are in Arithmetic Progression. The sum of the numbers representing A, C and E is 36 while that of A, C, E and G is 60. What is the sum of the numbers representing B, D, F and H?
(a) 96 (b) 66 (c) 72 (d) 84
1) Sum of three numbers in AP = 45
(a-d)+a+(a+d) = 45
3a = 45
a = 15
Sum of squares of numbers in AP = (a-d)^2+a^2+(a+d)^2 = 683
3a^2+2d^2 = 683
2d^2 = 683 - 675
d^2 = 4
d = +/- 2
Largest number among them = a+d = 15+12 = 17
2) (7)^1/p = (49)^1/q = (2401)^1/r
(7)^1/p = (7)^2/q = (7)^4/r
1/p = 2/q = 4/r
q = 2p and r = 2q
So P,q and r are in G.P with common ratio 2.
3) A+C+E = 36
a+a+2d+a+4d = 36
3(a+2d) = 36
a+2d = 12 -----> I
A+C+E+G = 60
So we get G = 60 - 36 = 24
So a+6d = 24 -----> II
By solving I and II we get
a= 6 and d = 3
So B+D+F+H = 4a + 16d = 24+48 = 72
Progressions SET-2 :-
1) There are two Arithmetic Progressions A and B such that their nth terms are given by An = 101 + 3(n 1) andBn = 150 + (n 1), where n is the set of natural numbers. The first 50 terms of A and B are written alternately i.e. A1B1A2B2..A50B50. What is the remainder when the number so formed is divided by 11?
(a) 0 (b) 1 (c) 9 (d) 10
2) Given that A > B > C and A60 = Bt = C120. If logA, logB and logC are in Arithmetic Progression, then what is the value of t?
(a) 40 (b) 60 (c) 80 (d) 120
3) In an increasing Arithmetic Progression, the product of the 5th term and the 6th term is 300. When the 9th term of this A.P. is divided by the 5th term, the quotient is 5 and the remainder is 4. What is the first term of the A.P.?
(a) 12 (b) 40 (c) 16 (d) 5
Progressions SET-2 :-
1) There are two Arithmetic Progressions A and B such that their nth terms are given by An = 101 + 3(n 1) andBn = 150 + (n 1), where n is the set of natural numbers. The first 50 terms of A and B are written alternately i.e. A1B1A2B2..A50B50. What is the remainder when the number so formed is divided by 11?
(a) 0 (b) 1 (c) 9 (d) 10
2) Given that A > B > C and A60 = Bt = C120. If logA, logB and logC are in Arithmetic Progression, then what is the value of t?
(a) 40 (b) 60 (c) 80 (d) 120
3) In an increasing Arithmetic Progression, the product of the 5th term and the 6th term is 300. When the 9th term of this A.P. is divided by the 5th term, the quotient is 5 and the remainder is 4. What is the first term of the A.P.?
(a) 12 (b) 40 (c) 16 (d) 5
1) The number so formed will be 101150......248199
We can write this number as
101*10^207+150*10^294+......+248*10^3+199*10^0
When 10^n is divided by 11, the remainder is 1 if n is even and -1 if n is odd.
Thus the remainder when the number is divided by 11 is
= 101+150-104+151-.....-248+199
-(101+104+.....+24
+(150+151+.....199)-(101+24
/2 * 50 + (150+199)/2 * 50-349/50 + 349/50 = 0
2) Given A>B>C,
let A^60 = B^t = C^120 = k
A = k^1/60
B= k^1/t
C=k^1/120
given A,B,C are in AP.. So A+C= 2B
logA+logC = 2 logB
log AC = 2 log B
log k^3/120 = 2 log K^1/t
3/120 log k = 2/t log k
3/120 = 2/t
t = 40*2 = 80
3) Let the 5th term of the AP be 'a' and common difference be 'd'.
The 6th term will be a+d and 9th term will be a+4d
a(a+d) = 300
5a+4 = a+4d -----> I
d = a+1 ----> II
From I and II we get
a = 12 or -25/2
If a=-25/2 then d will also be negative which is not possible in this AP.
Therefore a=12 and d=13
First term = a-4d = 12-52 = -40
Progressions SET-3 :-
1) If a positive integer n is subtracted from the squares of three consecutive terms of an Arithmetic Progression, the numbers obtained are 120, 232 and 376 respectively. What is the sum of the digits of n?
(a) 3 (b) 5 (c) 6 (d) 10
2) Given below is the sequence of all the natural numbers which dont contain the digit 0.
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13
What is the 200th term of this sequence?
(a) 242 (b) 241 (c) 255 (d) 253
3) What is the number of common terms in the two sequences given below?
S1 4, 10, 16, 22, ., 562.
S2 3, 8, 13, 18, ., 573.
(a) 20 (b) 40 (c) 36 (d) 18
in my final year engg now! hav given CAT 2011- 70 %ile 😞 (was not ready)
very confused if i should actually try for top 10 B-SCHOOLS..
10th- 89%
12th- 90.6%
b.e- 65 % 😞 any light of hope on my B.E % and getting calls from top B'S ?? or should i stop dreaming?
actually i had lots of fun in my b.e and in my final year got to know about how imp is grad % for a reputed MBA.. 😞 do i stand a chance now?? thnx a ton!
well your 10 th and 12 th score are good ... bit low on grad score ... but you have good chances of getting call from good colleges .. and for IIM's you donot know what criteria they are going to keep it can change .. according to last year criteria you have a clear chance of getting call from IIM-C and many new IIM's .... 😃
Progressions SET-3 :-
1) If a positive integer n is subtracted from the squares of three consecutive terms of an Arithmetic Progression, the numbers obtained are 120, 232 and 376 respectively. What is the sum of the digits of n?
(a) 3 (b) 5 (c) 6 (d) 10
2) Given below is the sequence of all the natural numbers which dont contain the digit 0.
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13
What is the 200th term of this sequence?
(a) 242 (b) 241 (c) 255 (d) 253
3) What is the number of common terms in the two sequences given below?
S1 4, 10, 16, 22, ., 562.
S2 3, 8, 13, 18, ., 573.
(a) 20 (b) 40 (c) 36 (d) 18
1) Let the three terms be a-d,a and a+d
120+n = (a-d)^2 ----> I
232+n = a^2 ------> II
376 + n = (a-d)^2 -----> III
Subtracting I from II and II from III, we get
2ad-d^2 = 112
2ad+d^2 = 144
Therefore d=4 and a=16
From I
120+n = 12^2
n = 24
Sum of digits of n = 2+4 = 6
2) In the first 200 natural numbers there are 20 numbers whose units digit is '0'.
There are 11 numbers whose tens digit is '0'.
There are 2 natural numbers whose both units and tens digit are '0'.
So in all there are 200-(20+11-2) = 171 such numbers which don't have '0' as their digit.
The 172nd,181st,190th and 199th terms of given series are 211,221,231 and 241 respectively.
So the 200th term is 242.
3) The first common term of two sequences is 28. As the common difference of two sequences are 6 and 5 respectively. We can say any number of the form '30n-2' where n is a natural number will be common to both the sequences. The maximum possible value of such common term will be
30*18-2 = 540-2 = 538
Hence number of common terms in S1 and S2 is 18.
Progressions SET-4 :-
1) In an arithmetic progression, if the ratio of (r+2)th term and the (r+5)th term is (r+3):(r+6) and the sum of the first 4r terms and sum of the first 7r terms are in the ratio 1:3. Find value of r.
a) 15 b) 18 c) 27 d) 29
2) Consider 20 infinite geometric progressions, whose first terms are 2,3,4,....21 respectively and the common ratios are 1/3,1/4,1/5....1/22 respectively. If s1,s2,s3....s20 denote the sums of these 20 geometric progressions. Find s1+s2+s3+...+s20.
a) 80 b) 320 c) 160 d) 250
3) Find 28383rd term in series 1234567891011..
a) 6 b) 7 c) 3 d) 4
1) In an arithmetic progression, if the ratio of (r+2)th term and the (r+5)th term is (r+3):(r+6) and the sum of the first 4r terms and sum of the first 7r terms are in the ratio 1:3. Find value of r.
a) 15 b) 18 c) 27 d) 29
2) Consider 20 infinite geometric progressions, whose first terms are 2,3,4,....21 respectively and the common ratios are 1/3,1/4,1/5....1/22 respectively. If s1,s2,s3....s20 denote the sums of these 20 geometric progressions. Find s1+s2+s3+...+s20.
a) 80 b) 320 c) 160 d) 250
3) Find 28383rd term in series 1234567891011..
a) 6 b) 7 c) 3 d) 4
Progressions SET-4 :-
1) In an arithmetic progression, if the ratio of (r+2)th term and the (r+5)th term is (r+3):(r+6) and the sum of the first 4r terms and sum of the first 7r terms are in the ratio 1:3. Find value of r.
a) 15 b) 18 c) 27 d) 29
2) Consider 20 infinite geometric progressions, whose first terms are 2,3,4,....21 respectively and the common ratios are 1/3,1/4,1/5....1/22 respectively. If s1,s2,s3....s20 denote the sums of these 20 geometric progressions. Find s1+s2+s3+...+s20.
a) 80 b) 320 c) 160 d) 250
3) Find 28383rd term in series 1234567891011..
a) 6 b) 7 c) 3 d) 4
1) [a+(r+2-1)d]/[a+(r+5-1)d] = (r+3)/(r+6)
By solving this we get
a=2d ----> I
Given that
(4r/2 * [ 2a+(4r-1)d] ) / (7r/2* [2a+(7r-1)d] = 1/3
Substitute 2d in place of a
4/7 * [(4rd+3d)/(7rd+3d)] = 1/3
48r+36 = 49r+21
r = 36-21 = 15
2) Sum of infinite GP = a/(1-r)
s1 = 2/(1-1/3) = 3
s2 = 3/(1-1/4) = 4
s3 = 4/(1-1/5) = 5
.
.
s20 = 21/(1-1/22) = 22
s1+s2+...+s20 = 20*(3+22)/2 = 10*25 = 250
3) 9+180+2700
=2889
25494 /4
remainder is 2
number till 28381 term
6373+999=7372
so next 2 digits from left 73
so the 28383rd number is 3
Progressions SET-5 :-
1) M and N are natural numbers such that by M = (5N 4) (5N + 1).
If 1N200,what is the harmonic mean of all the possible values of M?
a) 200^2/1001
b) 1001
c) 1001/200^2
d) 1/1001
2) A sequence 192, 360, 576. is formed by multiplying the corresponding terms of two different Arithmetic Progressions. What is the eighth term of the sequence?
(a) 2376 (b) 1040 (c) 2116 (d) None of these
3) From the first 20 natural numbers how many Arithmetic Progressions of five terms can be formed such that the common difference is a factor of the fifth term?
(a) 16 (b) 22 (c) 25 (d) 26
1) M and N are natural numbers such that by M = (5N 4) (5N + 1).
If 1N200,what is the harmonic mean of all the possible values of M?
a) 200^2/1001
b) 1001
c) 1001/200^2
d) 1/1001
2) A sequence 192, 360, 576. is formed by multiplying the corresponding terms of two different Arithmetic Progressions. What is the eighth term of the sequence?
(a) 2376 (b) 1040 (c) 2116 (d) None of these
3) From the first 20 natural numbers how many Arithmetic Progressions of five terms can be formed such that the common difference is a factor of the fifth term?
(a) 16 (b) 22 (c) 25 (d) 26
aimcats are going to start from 25 th may ...
1) Sum of three numbers in AP = 45
(a-d)+a+(a+d) = 45
3a = 45
a = 15
Sum of squares of numbers in AP = (a-d)^2+a^2+(a+d)^2 = 683
3a^2+2d^2 = 683
2d^2 = 683 - 675
d^2 = 4
d = +/- 2
Largest number among them = a+d = 15+12 = 17
here the answer is 16 i think.
as the three numbers are 14,15,16 with common difference 1
3) 9+180+2700
=2889
25494 /4
remainder is 2
number till 28381 term
6373+999=7372
so next 2 digits from left 73
so the 28383rd number is 3
hey can u pls explain this model of problem in detail
here the answer is 16 i think.
as the three numbers are 14,15,16 with common difference 1
No answer is 17. Just see the solution clearly...!!! Common difference will not be 1. It is 2.
vikranth504 Sayshey can u pls explain this model of problem in detail
What didn't you understand in this problem...???
hi folks i have recently started my preparation for cat
can some one post a simple way of finding the units digit of:
1^1*2^2*3^3*4^4*5^5*..........*100^100:shocked::shocked:
Progressions SET-5 :-
1) M and N are natural numbers such that by M = (5N 4) (5N + 1).
If 1N200,what is the harmonic mean of all the possible values of M?
a) 200^2/1001
b) 1001
c) 1001/200^2
d) 1/1001
2) A sequence 192, 360, 576. is formed by multiplying the corresponding terms of two different Arithmetic Progressions. What is the eighth term of the sequence?
(a) 2376 (b) 1040 (c) 2116 (d) None of these
3) From the first 20 natural numbers how many Arithmetic Progressions of five terms can be formed such that the common difference is a factor of the fifth term?
(a) 16 (b) 22 (c) 25 (d) 26
1) M=(5N-4)(5N+1) , 1When N=1, M = 1*6
When N=2, M = 6*11
When N=200, M = 996*1001
Harmonic Mean = 200/(1/5+1/5+....+1/5)
= 200/(1/5) = 1001
2) The nth term of an AP can be written in the form of a linear equation an+b.
So the product of 2 A.P's will have terms of the form pn^2+qn+r where n=0,1,2....
Putting n=0,1,2 we get r=192, p+q+r = 360 and 4p+2q+r = 576
Solving the above equations we get p=24 and q=144
Putting n=7 and values of p,q,r we get 8th term as 2376.
3) Led d be the common difference and a be the first term of AP.
The fifth term of the series will be a+4d. If a+4d is divisible by d then a should also be divisible by d. Hence the cases are
d=1 , a=1,2,3....16
d=2, a=2,4,6....16
d=3, a=3,6
d=4, a=4
so answer is 16+6+2+1 = 25
hi folks i have recently started my preparation for cat
can some one post a simple way of finding the units digit of:
1^1*2^2*3^3*4^4*5^5*..........*100^100
Units digit will be Zero..!! Because it is getting multiplied by numbers which have their units digit as zero...!!!
hi guys,
please let me know which topics you are discussing ..
Regards
kapil
hi folks i have recently started my preparation for cat
can some one post a simple way of finding the units digit of:
1^1*2^2*3^3*4^4*5^5*..........*100^100
The best way to find the unit's digit is to find the remainder of a number when it's divided by 10. So, for any x!, x>=10, this will always be zero!!!