1An = nXn + n(n+1)/2 = [n2 + n2(1 + 1/n)]/n
lim n-> inf An/ n = 3/2
Reason-Assertion Type prob
a)st-1 is true.st-2 is false
b)st-1 is false,st-2 is true
c)st-1 is true,st-2 is true,st-2 is correct explaination of st-1
d)st-1 is true,st-2 is true,st-2 is not correct explaination of st-1
Q1)St-1:Let N be the number of 3-digit numbers with distinct digits so that the digits in any number neither increase nor decrease.Then the sum of the divisors of N is 1055.
St-2:If p1,p2,p3 are distinct primes,then the sum of divisors of
N =p1α1 . p2α2 . p3α3 is
(p1α1+1 -1)(p2α2+1-1)( p3α3+1-1)/(p1-1)(p2-1)(p3-1)
Q2)St-1:There are 5 integers between 1 and 200, each of which has ten divisors.
St-2:If p1,p2,p3 are distinct primes,then number of divisors of p1α1 . p2α2 . p3α3 is (1+α1)(1+α2)(1+α3).
Paragraph based
Let n ε N.The A.M.,G.M.,H.M. and R.M.S.(root mean square) of the n numbers n+1,n+2,n+3,......n+n are An,Gn,Hn,Rn respectively.Then
Q1)limn →∞An/n=
a)1
b)3/2
c)2
d)1/2
Q2)limn →∞Gn/n=
a)1/e
b)2/e
c)3/e
d)4/e
Q3)limn →∞Hn/n=
a)ln2
b)1/ln2
c)1
d)e
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14 Answers
1357y=Gn/n=[(n+1)(n+2)...(n+n)]1/n/n=[(1+1/n)(1+2/n)...(1+n/n)]1/n
logy=(1/n) Σ(r=1 to n) log(1+r/n)
logy=0∫1log(1+x)dx=2log2-1=log4-loge=log(4/e)
y=4/e
1another tukka method :
lim n-> inf An/n = 1.5
AT INF , am -> gm
so, Gn/n = 4/e ≈ 1.47 ≈ 1.5 [5]
1sorry i am out of perm-combi and probability ....
bhaiya(s) or someone else will help ...
1again for q.3 >> ans is B.
ln2 = 0.693
1/ln2 = 1.44 approx
1 = 1 [3]
e = 2.23 ...blah blah
**** AM=GM=HM only wen at the limitimg case each term tends to each other..
here wen u put n=inf, all terms tends to 1 .
1the word tukka* is wrong here [3]
it is logical i think ... just to get ans fast :)
1357St-2:If p1,p2,p3 are distinct primes,then number of divisors of p1α1 . p2α2 . p3α3 is (1+α1)(1+α2)(1+α3).
statement 2 is true
for 10=5*2 divisors
we have (1+α1)(1+α2)=2*5
so α1=1 and α2=4
then N=21.34 or 24.31 or 24.51 or 24.71 or 24.111
so 5 N's possible
62Q1)St-1:Let N be the number of 3-digit numbers with distinct digits so that the digits in any number neither increase nor decrease.Then the sum of the divisors of N is 1055.
St-2:If p1,p2,p3 are distinct primes,then the sum of divisors of
N =p1α1 . p2α2 . p3α3 is
(p1α1+1 -1)(p2α2+1-1)( p3α3+1-1)/(p1-1)(p2-1)(p3-1)
N be the number of 3-digit numbers with distinct digits so that the digits in any number neither increase nor decrease.
no of such numbers is 10C3x(3!-2) = 10x3x8x2 = 3x5x25
no of ways = 35x4x6 = 140x6 = 840
I Dont think that I am missing something here.. but ...
Statement 2 is true..
621055 = 5x211 = ??!!!
I dont think that 5 can be written as 1+p+...
so 1055 has to itself be wrtten in that form
1+5+25+125+625 does not work
1+7+49+343+ does not work
1+11+121+.. does not work
1+13+169...
so it has to be of the form p(1+p) = 1055-1
p(1+p) = 1054 which again does not seem to have an integer solution!
1357Q3)limn →∞Hn/n=
1/Hn=[1/(n+1) + (1/(n+2) + 1/(n+3) + ...]/n
z=n/Hn=1/(n+1) + 1/(n+2)....
z=(1/n)Σ(r=1 to r=n) 1/(1+r/n)
z(lim n→∞)=0∫1dx/(1+x)=ln2
so 1/z=1/ln2