@geetlove said: Q) last two digits (201 x 202 x 203 x 204 x 246 x 247 x 248 x 249)^2
pl provide systematic solution that is by dividing 100
instead of multiplying 01*02*03*04*46*47*48*49 an easier alternative is to first find out the squares of the last two digits of each of the numbers by writing them as (200+1)^2 (200+2)^2 (200+3)^2 (200+4)^2 (250-4)^2 (250-3)^2 (250-2)^2 (250-1)^2. now the last two digits become 01*04*09*16*16*09*04*01 or rather 256*81*16.. hence the last two digits are 76.. the calculation becomes faster.. :)
the given series can be written as 1/4 * ( 1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ...... )
all terms after 1/7 are cancelled so what remains is 1/4* 1/3 = 1/12
answer must be 1/12 just check and clarify
cheers
heyy, thanks for the help.. thanks a lot!! but my doubt is that the second term of the 1st 1/7 and the second 1/7 are getting canceled. so the 2nd part of the 1st term and the 1st part of the second term are getting canceled. that way will not the last term have a part of it remaining???? as in the last term will be 1/x-1/y, now the 1/x will get cancelled eventually coz the previous term will have a negative version of that, but what will happen to the 1/y? how do we cancel that? please if you cud elaborate on that a bit., rest is all fine.. ya, the ans is 1/12.. :).. thanks again for your help mate..!!
take x=1 y=2 z=1 so substitute in given expression u get 12
whereas x=0 y=3 z=0 you get 18
myt be a better way of finding out then let me kno ! i wud rather substitute and eliminate options
rest of the questions are not clear ! check them out thanks
if we take x=1, y=2 and z=1 then the answer is coming to 10. 1^2 + 2*2^2 + 1^2 = 1+8+1= 10. please clarify that bit.. n well the last question is 243^log x base 81- 2x= 2^log(x+2) base 16 -8. they have asked to find the sum of all real values that satisfy the above equation.
heyy, thanks for the help.. thanks a lot!! but my doubt is that the second term of the 1st 1/7 and the second 1/7 are getting canceled. so the 2nd part of the 1st term and the 1st part of the second term are getting canceled. that way will not the last term have a part of it remaining???? as in the last term will be 1/x-1/y, now the 1/x will get cancelled eventually coz the previous term will have a negative version of that, but what will happen to the 1/y? how do we cancel that? please if you cud elaborate on that a bit., rest is all fine.. ya, the ans is 1/12.. .. thanks again for your help mate..!!
see after 1/3, every positive term has its negative version till infinity--- all will get cancelled 😃 1/3- 1/7+1/7-1/11+1/11-1/15...........-1/x+1/x-..................... -inf+inf
see after 1/3, every positive term has its negative version till infinity--- all will get cancelled 1/3- 1/7+1/7-1/11+1/11-1/15...........-1/x+1/x-..................... -inf+inf
oh ya.... I get it!! fine.. thanks a tonn.. 😃 :).. just got confused in the last step.. :).. thanks again!!
and for makin this thing minimum make all the terms of am equal.. ie x^2 = y^2 = z^2 or x=y=z; put this in eqn 4x= -6 implying x= -6/4 now put hthis value in given expression u'l get the min valu. but it is coming to be 13.5
1. find the sum of the series- 1/(3*7)+ 1/(7*11)+ 1/(11*15)+........ answer is 1/3..
2. if x, y, z are any real numbers then find the minimum possible value of x^2+ 2y^2+ z^2+ 2yz subject to x+2y+z =-6.. answer is 12
3. 16
A= [{2n^3 + 2n^2 +n^1/2(n+3)-3}/ {4n*n^1/2(n^1/2+3)+9}] then n necessarily satisfies which of the following? ans is 0.508
4. find the sum of all real values of 'x' that satisfy the equation-
(243^logx base 81)- 2x= (2^logx+2 base 16)-8... answer is 20. please kindly help me with these sums.. I want to solve them in the easiest way possible. TIA.. .. happy learning..!!
in responce to ur last question: lhs ccan be simplified to 243^(1/4) - 2x and rhs is (x + 2)^(1/4) -8
for the sum of roots of this equation make it a polynomial equation of degree 4 and sum of roots will be given by coefficient of x^3 the process is lenthy but will be reduced too much if we concentrate ourself only for the coefficient of x cube neglecting other terms.. and when i followed,, it was found to be negative and was not an integer..
how many no.s are there btw 100 and 1000 which have exactly one of their digit as 7 ?
(ans-225). please reply with the elaborate solution...how to calculate??
when 7 is at hundredth place, we can have 9 digits at tens and 9 at ones.. total 81 such numbers are possible. now when 7 is at tens then 8 are likely to be at hundredth, and 9 at ones, giving out 72 results.. similarly when 7 is at ones then 8 at hundredth and 9 at ones are possible giving 72 such numbers,, adding all of them gives 225 such numbers..
How many 3 digit even numbers are there such that if they are subtracted from their reverse the result is positive multiple of 66 ?
the answer is 10; . let n= 100x + 10y + z and m = 100z + 10y + x; then n-m = 99(x - Z), and for the number to be divisible by 66 we jus need 2 as a factor . now since the number is even then for z = 0, possible values x can take are.. 2,4,6,8; for z=2, x = 4,6,8; for z=4, x = 6,8; and for z=6, x = 8; z cannot be more than 6, cz then we'll obtain a negative quotient... so total such numbers are 10..
