Thank u
n n
C = C
8 2
Find
n
C
2
C = C
8 2
Find
n
C
2
@Logrhythm said:(1000-1)*abc = 132*def1000abc = 132*def + abcso a=8,b=6 and c=8=> 868000 - 868 = def*132=> def = 867...so a+b+c+d+e+f = 8+6+8+8+6+7 = 43...
How u arrived at this. a=8,b=6 and c=8
How will you measure exact 250 ml if all u have is a cylindrical measuring tube of 1L and no other measuring tool..???
there are four distinct numbers such that first and last ones are equal.the first three are in A.P and last three are on G.P. find the common ratio of last three
1
-1
1/2
-2
@raopradeep said:there are four distinct numbers such that first and last ones are equal.the first three are in A.P and last three are on G.P. find the common ratio of last three1-11/2-2
1?
@raopradeep said:@paridhi11890OA is -2
let the nos be
a, a+d, a+2d, mr^2
given a=mr^2
also a+d=m
a+2d= mr
then
d= m(r-1)
so mr^2+m(r-1)= m
r^2+r-1=1
r(r+1)=2
r=1 satisfies this..
Wat is wrong in my method?????? :(
@raopradeep said:how to approach tell the method
How will you measure exact 250 ml if all u have is a cylindrical measuring tube of 1L and no other measuring tool..???
@raopradeep
let the no be a b c d
now 1st and last no are same, hence , (a=d).. so the sequence becomes a b c a
Now 1st 3 are in AP, hence a, a+d , a+2d
and last 3 are in GP ,Hence (last 3 nos are) a+d , a+2d , a
Now apply for a GP (b^2=ac)..
Hence (a+d)^2 = ax(a+2d)
On solving we get either d=0 or d= -3a/4
so on subsituting the value of d in the sequence we get
a , a/4 ,-a/2 ,a
hence r=-2
now 1st and last no are same, hence , (a=d).. so the sequence becomes a b c a
Now 1st 3 are in AP, hence a, a+d , a+2d
and last 3 are in GP ,Hence (last 3 nos are) a+d , a+2d , a
Now apply for a GP (b^2=ac)..
Hence (a+d)^2 = ax(a+2d)
On solving we get either d=0 or d= -3a/4
so on subsituting the value of d in the sequence we get
a , a/4 ,-a/2 ,a
hence r=-2
@ani4588 said:@raopradeeplet the no be a b c dnow 1st and last no are same, hence , (a=d).. so the sequence becomes a b c aNow 1st 3 are in AP, hence a, a+d , a+2dand last 3 are in GP ,Hence (last 3 nos are) a+d , a+2d , aNow apply for a GP (b^2=ac).. Hence (a+d)^2 = ax(a+2d)On solving we get either d=0 or d= -3a/4so on subsituting the value of d in the sequence we geta , a/4 ,-a/2 ,ahence r=-2
Please tell what is wrong in my method?
@raopradeep said:there are four distinct numbers such that first and last ones are equal.the first three are in A.P and last three are on G.P. find the common ratio of last three1-11/2-2
let the numbers be a-d, a, a+d, a-d
=> a/(a+d)=(a+d)/(a-d)
=> a^2-a*d=a^2+2*a*d+d^2
=> d^2=-3*a*d
=> d=-3*a
==> common ratio = (a+d)/a = (a-3*a)/a = -2
Okay... I see that my equation satifies... r=-2 as well :banghead:
How did I ignore it... :mad:
@paridhi11890 said:let the nos bea, a+d, a+2d, mr^2given a=mr^2also a+d=ma+2d= mrthend= m(r-1)so mr^2+m(r-1)= mr^2+r-1=1r(r+1)=2r=1 satisfies this..Wat is wrong in my method??????
r*(r+1) = 2
=> r^2+r-2 =0
=> (r-1)(r+2) = 0
=> r=1 or -2
:neutral:
@vijay_chandola said:r*(r+1) = 2=> r^2+r-2 =0=> (r-1)(r+2) = 0=> r=1 or -2
Yeah...I know... How did i not look at -2 :facepalm: :|
@paridhi11890 said:Yeah...I know... How did i not look at -2 :facepalm:
chill , everybody is allowed to commit a few mistakes ....