that reminds me... will 0.123456789101112131... be a rational number since we always know what the next digit/no. is ?!
my take :
option d) both 1 and 2..
if correct then will post the approach...
:o
Hi ...
its correct..tell me the procedure
take the terms to be a, a+d, a+2d ................... a+19d.(if n = 20)
1st series will contain a, a+3d, a+6d........a+15d, a+18d
sum of the series 7a+63d
which is divisible by 7 and will give a+9d, which is part of our series..
similarly 2nd series sum = 7a+ 70d, which will give a+10d..satisfies the condition..
now check for 26 terms also you will see that it is also satisfying the condition..
this was my approach hope u got it...
:-P
progessions looks like a favort in CAT, ain't it?
Al d'Best Saysi suppose the best approach would be to take the first 2 or 3 terms and then look at the kind of result?
brother plz use the quote button to reply..as it will be much more clear that u are replying to which post...
divishth Says1972 is the Asnwer
pls explain this
akshbhardwaj Sayspls explain this
divishth Says1972 is the Asnwer
Please help me in solve this:
The series of numbers (1,1/2,1/3,....1/1972) is taken.Now 2 numbers are taken from this series say x,y.Then the operation x+y+x.y is performed to get a consolidated no.the process is repeated.what will be the value of the set after all the numbers are consolidated into one number.
a)1970
b)1971
c)1972
d)none
Take x = 1,y=1/2..Value = x+y+xy = 1+1/2 + 1/2 = 2
Take x=2 and y=1/3...Value = x+y+xy = 2+1/3 + 2/3 = 3
Take x=3 and y=1/4..value = 3+1/4 + 3/4 = 4
...Similarly last term would be 1971 + 1/1972 + 1971/1972 = 1972
Hi puys,
Can you give the solns for the follwing:::: These are from Arun sharma Progressions
1. Find the value of 1-2-3+2-3-4+.......+ upto 100 terms??
a. -694 b.-626 c.-624 d.-549 e. -676
2. If a man saves Rs.1000 each year and invests at the end of the year at 5% compound interest how much will the amount be at the end of 15 years?
a. 21,478 b. 21,578 c. 22,578 3. 22,478 e. 22,178
3. Find the infintie sum of the series 1+1/3+1/6+1/10+/15+......
a. 2 b.2.25 c. 3 d. 4
4. An arithmetic progression P consists of n terms. From the progression three different progressions P1P2 and P3 are created such that P1 is obtained by the 1st,4th,7th....... terms of P, P2 has 2nd ,5th,8th ....... terms of P, P3 has the 3rd,6th,9th.terms of P. It is found that of P1,P2,P3 two progressions have the property that their average itself is a term of the original progression P. Which of the following can be a possible value of n??
a. 20 b.26 c.36 d. Both 1 and 2
5 . Product of the fourth term and fifth term of an AP is 456. Divivsion of the ninth term by the fourth term of the progression gives quotient as 11 and remainder as 10. Find the first term of the AP
a. -52 b. -42. c. -56 d.-66 e.-50
6. Find the sum of series 1.2+2.2^2+3.2^3+........+100.2^100
7. If x,y,z are in GP and a^x,b^y and c^z are in ?
a. AP b. GP c. HP
8. If three +ve real numbers x,y,z are in AP such that xyz =4 ,then what will be the minimum value of y?
a. 2^1/3 b.2^2/3 c.2^1/4 d. 2^3/4 e. 2^4/3
Please give detiled solutions
1. Find the value of s=1-2-3+2-3-4+3-4-5+.......+ upto 100 terms??
S Is forming a pattern n-(n+1)-(n+2)
If we start from 1-2-3 it gives -4
2-3-4=-5------------------------------------till 33-34-35( 33*3=99 terms in all) in the end only 34 will be left
-4-5-6-7-------- -35+ 34
-(4+5+6+7------+35) +34
-(33(2*4+(33-1)1)/2+34
=-660+34=-626
Helloooooooo 
1. Find the value of s=1-2-3+2-3-4+3-4-5+.......+ upto 100 terms??
S Is forming a pattern n-(n+1)-(n+2)
If we start from 1-2-3 it gives -4
2-3-4=-5------------------------------------till 33-34-35( 33*3=99 terms in all) in the end only 34 will be left
-4-5-6-7-------- -35+ 34
-(4+5+6+7------+35) +34
-(33(2*4+(33-1)1)/2+34
=-660+34=-626
don't u think that the sum of (33-34-35) should be 36 and it should be like this -(4+5+6+7+.........+36)+34 bt ans will come out of the options.....bt still...........
what is the remainder when 128^1000 is divided by 153?
a)103
b)145
c)118
d)52
how to solve
how to solve
what is the remainder when 30^72^87 is divide by 11
a)5
b)9
c)6
d)3
N = 202 * 20002 * 200000002 * 20000000000000002 * 200000....2(31 zeroes) .The sum of the digits in this multiplication will be
a)112
b)160
c)144
d)cannot be determined
Please explain how to solve this
how to solve
what is the remainder when 30^72^87 is divide by 11
a)5
b)9
c)6
d)3
The Euler number of 11 is 10,
consider 72^87, the last digit of the expansion of 72^87 will be 8
so finally we get to find the remainder of
remainder when 30^8 is divided by 11
=> 8^8 divided by 11
=> 64*64*64*64* by 11
=>9*9*9*9 by 11
=> 81*81 by 11
=> 4*4 by 11
=> remainder = 5
what is the remainder when 128^1000 is divided by 153?
a)103
b)145
c)118
d)52
how to solve
how to solve
what is the remainder when 30^72^87 is divide by 11
a)5
b)9
c)6
d)3
N = 202 * 20002 * 200000002 * 20000000000000002 * 200000....2(31 zeroes) .The sum of the digits in this multiplication will be
a)112
b)160
c)144
d)cannot be determined
Please explain how to solve this
If you've more than one question, post all of them in one post and don't unnecessary make separate posts for separate questions. Puys can still answer individual questions from one post.
N = 202 * 20002 * 200000002 * 20000000000000002 * 200000....2(31 zeroes) .The sum of the digits in this multiplication will be
a)112
b)160
c)144
d)cannot be determined
Please explain how to solve this
2*2*2*2*2(101 * 10001 * 100000001 * 10000000000000001 * 1000...001(31times))
32 * (1010101 * * 100000001 * 10000000000000001 * 1000...001(31times))
So final Term Would be 32323232.....3232 (32 would be 32 times)
So Sum of digits = 3*32 + 2*32=160
what is the remainder when 128^1000 is divided by 153?
a)103
b)145
c)118
d)52
how to solve
E(153) = 96
128^1000 mod 153 = 128^40 mod 153
2^280 mod 153
2^280 mod 3 = 1
2^280 mod 17 = 2^8 mod 17 = 1
3a + 1 = 3b+1 = 17c+1
Remainder = 52
A three digit no. 'mnp' is a perfect square and the no. of factors it has is also a perfect square.It is also known that the digits m,n,p are distinct.
1)if (m+n+p) is also a perfect square,what is the no. of factors of the six digit no. mnpmnp?
a)32
b)72
c)48
d)cannot be determined
2)if the fourth power of the product of the digits of the no. mnp is not divisible by 5, what is the no. of factors of the 9 digit no,mnpmnpmnp?
a)32
b)72
c)48
d)cannot be determined
2)How many integer values of x and y satisty the exp 4x+3y=3 where xa)284
b)285
c)286
d)none
Please explain me the solution
A three digit no. 'mnp' is a perfect square and the no. of factors it has is also a perfect square.It is also known that the digits m,n,p are distinct.
1)if (m+n+p) is also a perfect square,what is the no. of factors of the six digit no. mnpmnp?
a)32
b)72
c)48
d)cannot be determined
2)if the fourth power of the product of the digits of the no. mnp is not divisible by 5, what is the no. of factors of the 9 digit no,mnpmnpmnp?
a)32
b)72
c)48
d)cannot be determined
Please explain me the solution
my take for both the questions is
CBD option d..
if correct will tell u the approach..
my take for both the questions is
CBD option d..
if correct will tell u the approach..
The ans for ist question is (d) and for the 2nd question is (c)
209598 SaysThe ans for ist question is (d) and for the 2nd question is (c)
for 2nd question also there are various values coming..
196, 441 and other also..
so hw can we determine a single value..
:shocked: