@dragster said:Q. C.I for 3 years = 1260 ; CI for 4 years = 1512 , compounded annually.principal??(don remember the answr.. but cudnt solve.. need aproach.. )
@dragster 1269 * 1.2 = 1512. hence interest is 20%. hence P=1730.77??
@krum said:C.I dia hai, r u sure b'day boy , happy b'day
bday on 14th bhai 😁 😁 pta nahi shayad lat de rakha hai on PG yaa made a mistake amount le liya tha
@piyushrohella12 said:form two eqn divide (1+r/100)=1512/1260=1.2P*(1.2)^3=1260p=729.16 feels like m back home sabko bhai ka ____/\____
@piyushrohella12 the interest is 1260 and not the amount .
@piyushrohella12 said:form two eqn divide (1+r/100)=1512/1260=1.2P*(1.2)^3=1260p=729.16 feels like m back home sabko bhai ka ____/\____
o bhai thek se pad questn..
itna sa hota toh me post karta..
itna sa hota toh me post karta..@dragster said:o bhai thek se pad questn.. itna sa hota toh me post karta..
o bhai upar dekho galti pakad lii :P
@dragster said:Q. C.I for 3 years = 1260 ; CI for 4 years = 1512 , compounded annually.principal??(don remember the answr.. but cudnt solve.. need aproach.. )
are u sure, 3 and 4 years hai rather than 3rd and 4th year?
@krum said:are u sure, 3 and 4 years hai rather than 3rd and 4th year?
mistake.. precisely.. 3rd n 4th year.. thek point kiya.
now try.. may b ho jaaye..
now try.. may b ho jaaye..
@dragster said:mistake.. precisely.. 3rd n 4th year.. thek point kiya.now try.. may b ho jaaye..
p(1+r/100)^4-p(1+r/100)^3=1512
p(1+r/100)^3-p(1+r/100)^2=1260
dividing
==> ((1+r/100)^4-(1+r/100)^3)/( (1+r/100)^3-(1+r/100)^2)=1512/1260
==>(1+r/100)=6/5
==>r=20
p(1.2^3-1.2^2)=1260
=>p=1260/1.2^2(.2)= 4375
p(1+r/100)^3-p(1+r/100)^2=1260
dividing
==> ((1+r/100)^4-(1+r/100)^3)/( (1+r/100)^3-(1+r/100)^2)=1512/1260
==>(1+r/100)=6/5
==>r=20
p(1.2^3-1.2^2)=1260
=>p=1260/1.2^2(.2)= 4375
@dragster said:mistake.. precisely.. 3rd n 4th year.. thek point kiya.now try.. may b ho jaaye..
(1 + r/100) = R
P(R^3 - R^2) = 1260
P(R^4 - R^3) = 1512
=> Divide => R = 1.2
P*(R^2)(R-1) = 1260 => P = 1260*5/(1.44) = 4375
@dragster said:Q. C.I for 3rd year = 1260 ; CI for 4th year = 1512 , compounded annually.principal??(don remember the answr.. but cudnt solve.. need aproach.. )
it shud be 4375?
1260 *r/100= (1512-1260)
r=20
p(1.2^3-1.2^2)=1260
p=4375
correct me if wrng
@krum said:p(1+r/100)^4-p(1+r/100)^3=1512 p(1+r/100)^3-p(1+r/100)^2=1260dividing==> ((1+r/100)^4-(1+r/100)^3)/( (1+r/100)^3-(1+r/100)^2)=1512/1260==>(1+r/100)=6/5==>r=20p(1.2^3-1.2^2)=1260=>p=1260/1.2^2(.2)= 4375
gud.. canceling this was the catch.. came out easy 
What's up
@maddy2807 said:it shud be 4375?1260 *r/100= (1512-1260)r=20p(1.2^3-1.2^2)=1260p=4375correct me if wrng
right..
a(n) is a sequence of +ve integers such that a(n-1)*a(n+1)=a(n)^2+1 for all n>=2. If a(2)=5 and a(5)=89, Find a(7).?
1,233,434,610
@pankaj1988 said:a(n) is a sequence of +ve integers such that a(n-1)*a(n+1)=a(n)^2+1 for all n>=2. If a(2)=5 and a(5)=89, Find a(7).?1,233,434,610
610
@pankaj1988 said:a(n) is a sequence of +ve integers such that a(n-1)*a(n+1)=a(n)^2+1 for all n>=2. If a(2)=5 and a(5)=89, Find a(7).?1,233,434,610
a1.a3 = a2 ^2 +1
a1.a3 = 26
a3 =13
now you can calculate a4 =34
a6 = 233
and therefore a(7) comes to be 610
some other approach?
@pankaj1988 said:a(n) is a sequence of +ve integers such that a(n-1)*a(n+1)=a(n)^2+1 for all n>=2. If a(2)=5 and a(5)=89, Find a(7).?1,233,434,610
(a3)^2+1=5*a4
(a4)^2+1=a3*89
=>a3=13, a4=34
(a5)^2+1=34*a6
=>a6=(89^2+1)/34=233
(a6)^2+1=89*a7
=>a7=(233^2+1)/89
=>a7=610
@pankaj1988 koi easy approach ho to do post
(a4)^2+1=a3*89
=>a3=13, a4=34
(a5)^2+1=34*a6
=>a6=(89^2+1)/34=233
(a6)^2+1=89*a7
=>a7=(233^2+1)/89
=>a7=610
@pankaj1988 koi easy approach ho to do post
@milestogo3 said:a1.a3 = a2 ^2 +1a1.a3 = 26a3 =13now you can calculate a4 =34a6 = 233and therefore a(7) comes to be 610some other approach?
@krum said:(a3)^2+1=5*a4(a4)^2+1=a3*89=>a3=13, a4=34(a5)^2+1=34*a6=>a6=(89^2+1)/34=233(a6)^2+1=89*a7=>a7=(233^2+1)/89=>a7=610@pankaj1988 koi easy approach ho to do post
mere paas bhi koi easy approach nhi hai...maine khud isi liye post kiya tha
gn...
@pankaj1988 said:mere paas bhi koi easy approach nhi hai...maine khud isi liye post kiya thagn...
arey mein function ke hisaab se koi pattern dhooond rha tha...
just had an 'apple' and got it. :mg:
next term(n) = (n-1) * 3 - (n-2)