@anytomdickandhary said:Hey can you please elaborate on your question?you can check that (n+2)*(n+3)^2 gives the terms of the sequence given for n=1,2,3....Not very sure if I understand your question here.ATDH.
Sir how did you got the equation..?
@ScareCrow28 said:@vijay_chandola@anytomdickandhary sir.. your general terms are not matching in order of "x" Why so?? I couldn't understand..
Edited my previous post. Made a little mistake 
Nth term=(n+2)*(n+3)^2 (as sir has derived)
First term= 3*4^2=48
second term=4*5^2=100
third term=5*6^2=180
First term= 3*4^2=48
second term=4*5^2=100
third term=5*6^2=180
fourth term =6*7^2=294 and so o...
Seems correct to me :)
@vijay_chandola said:Edited my previous post. Made a little mistake Nth term=(n+2)*(n+3)^2 (as sir has derived)First term= 3*4^2=48second term=4*5^2=100third term=5*6^2=180fourth term =6*7^2=294 and so o...Seems correct to me
I understand now! 😃 Thanks for editing your post! :P
@anytomdickandhary Confusion cleared Sir!
@maneeshp said:how you got this? Nth term=(n+2)*(n+3)^2See while solving , when x= 1, a +b+c+d = 28,when x =2,8a+4b+2c+d = 34,when x= 3,27a+9b+3c+d =40u got the above equation after solving these 3 equations? correct me if i am wrong.
no..the equations are to be equated to the original terms 48,100,180..
48, 100, 180, 294, 448...
This process is extremely mechanical..... it looks long because I have explained steps in detail. Please dont get intimidated by the length of the post. Its an iterative process that I have explained.
Rule1: if the successive difference becomes constant after 'n' iteration then the general term is of order 'n'
-------------48---------- 100----------180------------294----------448-------
1st Diff-----------52------------80------------114------------154-----------
2nd Diff---------------------28--------------34----------------40------------------
3rd Diff---------------------------6-----------------6---------------------
So difference becomes constant in 3rd level so the general term must be of order 3
Rule2: co-efficient of n^k is given by (constant diff)/k!. i.e in our case constant difference = 6 (as derived above) hence co-efficient = 6/3! = 1
so we know that the general term must be n^3 + an^2 +bn + c. This means that if we remove n^3 from each term we are left we quadratic expression.
i.e. (48-1^3), (100-2^3), (180-3^3),
i.e 47 , 92 , 153, 230
---------------47----------------92-------------------153-----------
1st diff--------------45-----------------61--------------
2nd diff---------------------16-----------------
applying rule2: we get coeff of n^2 as (16/2!) = 8
Now we remove the 8n^2 from the series to get the linear part.
(47-8*1^2), (92- 8*2^2),
----------39------------60-----
1st diff---------21---------
applying rule2: we get coeff of n^1 as (21/1!) = 21
now remove the 21n from series to get the constant term
=39-21*1 = 18
hence the series is n^3 + 8n^2 + 21n + 18
I like this process because it is very mechanical and if we need to check back then it is easier
ATDH.
@anytomdickandhary __/\__ :thumbsup:
Usually I do not spam on this thread, but the above post deserves one :P
That was one hell of a good concept! Thanks 😁 :D
@vijay_chandola said:@anytomdickandhary __/\__ Usually I do not spam on this thread, but the above post deserves one That was one hell of hell of a good concept! Thanks
If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4ďťż
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4ďťż
@Gaul said:If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?A. 1/4B. 1/3C. 1/2D. 2/3E. 3/4
1-120/360
=1-1/3
=2/3
=1-1/3
=2/3
@Gaul said:If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?A. 1/4B. 1/3C. 1/2D. 2/3E. 3/4ďťż
D. 2/3
probability that the length of the chord is greater than 2
= probability that the angle subtended by the arc at the centre of the circle is greater than 60
= 240/360
= 2/3
probability that the length of the chord is greater than 2
= probability that the angle subtended by the arc at the centre of the circle is greater than 60
= 240/360
= 2/3
@Gaul said:If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?A. 1/4B. 1/3C. 1/2D. 2/3E. 3/4ďťż
2/3?
@Gaul said:If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?A. 1/4B. 1/3C. 1/2D. 2/3E. 3/4
With a total arc 360-60-60=240 degree, required length would be greater than 2.
Hence, probability = 240/360=2/3
usually i dnt like any post but this one deserves kudos.. great concept
Moongerilal's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father, and Moongerilal died four years after his son.
How many years did Moongerilal live??
@vijay_chandola said:
@Anivesh90 said:
@grkkrg said:
@krum said:1-120/360=1-1/3=2/3
Spot on! yes, 2/3 it is.
@Gaul said:If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?A. 1/4B. 1/3C. 1/2D. 2/3E. 3/4
240/360=2/3....angle > 60
@bullseyes said:@anytomdickandharyusually i dnt like any post but this one deserves kudos.. great conceptMoongerilal's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father, and Moongerilal died four years after his son.How many years did Moongerilal live??
84?
@bullseyes said:@anytomdickandharyusually i dnt like any post but this one deserves kudos.. great conceptMoongerilal's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father, and Moongerilal died four years after his son.How many years did Moongerilal live??
x/6 + x/12 + x/7 + 5 + x/2 + 4 = x
==> x = 84
84 years
@bullseyes said:@anytomdickandharyusually i dnt like any post but this one deserves kudos.. great conceptMoongerilal's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father, and Moongerilal died four years after his son.How many years did Moongerilal live??
84
x/6 + x/12 + x/7 + 5 + x/2 + 4
= (14x + 7x + 12x)/84 + 9 + x/2
= 75x/84 + 9
= x
9x/84 = 9
x = 84
x/6 + x/12 + x/7 + 5 + x/2 + 4
= (14x + 7x + 12x)/84 + 9 + x/2
= 75x/84 + 9
= x
9x/84 = 9
x = 84
@bullseyes said:@anytomdickandharyusually i dnt like any post but this one deserves kudos.. great conceptMoongerilal's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father, and Moongerilal died four years after his son.How many years did Moongerilal live??
108?
@anytomdickandhary said:48, 100, 180, 294, 448...This process is extremely mechanical..... it looks long because I have explained steps in detail. Please dont get intimidated by the length of the post. Its an iterative process that I have explained.Rule1: if the successive difference becomes constant after 'n' iteration then the general term is of order 'n'-------------48---------- 100----------180------------294----------448-------1st Diff-----------52------------80------------114------------154-----------2nd Diff---------------------28--------------34----------------40------------------3rd Diff---------------------------6-----------------6---------------------So difference becomes constant in 3rd level so the general term must be of order 3Rule2: co-efficient of n^k is given by (constant diff)/k!. i.e in our case constant difference = 6 (as derived above) hence co-efficient = 6/3! = 1so we know that the general term must be n^3 + an^2 +bn + c. This means that if we remove n^3 from each term we are left we quadratic expression. i.e. (48-1^3), (100-2^3), (180-3^3), i.e 47 , 92 , 153, 230---------------47----------------92-------------------153-----------1st diff--------------45-----------------61--------------2nd diff---------------------16-----------------applying rule2: we get coeff of n^2 as (16/2!) = 8Now we remove the 8n^2 from the series to get the linear part.(47-8*1^2), (92- 8*2^2), ----------39------------60-----1st diff---------21---------applying rule2: we get coeff of n^1 as (21/1!) = 21now remove the 21n from series to get the constant term=39-21*1 = 18hence the series is n^3 + 8n^2 + 21n + 18I like this process because it is very mechanical and if we need to check back then it is easierATDH.
I am extremely sorry for spamming here...but seriously elated on seeing this 😃 I won't mind getting banned for spamming though 😛 I have seldom come across concepts more beautiful than this..(More than 90% from ATDH Sir easily) Lucky to have seen this..
The post should get its appreciation and the person posting this should be given its due respect.
Thank a lot @anytomdickandharry Sir!!