Few students are standing one behind the other. They are asked to take...
Let the first student picks x chocolates.
∴ The total number of students = 2x
∴ The number o f chocolates tak en b y the students: x , x , 2x, 4x, 8x, 1 6 x ......
We observe that the second term onwards is a G.P.
Thus, the total number of students = 8
Hence, option 3.
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Few students are standing one behind the other. They are asked to take...
Problem Analysis:
Let's assume the number of students in the column is 'n'.
The first student takes as many chocolates as half the total number of students in the column, which means the first student takes n/2 chocolates.
The second student takes twice the number of chocolates taken by the student just before him/her, which means the second student takes 2 * (n/2) = n chocolates.
The third student takes twice the number of chocolates taken by the student just before him/her, which means the third student takes 2 * n = 2n chocolates.
And so on...
Approach:
To find the number of students, we need to sum up the total number of chocolates picked by all the students, which is given as 512.
Let's consider the sum of chocolates picked by the first k students.
The first student picks n/2 chocolates.
The second student picks n chocolates.
The third student picks 2n chocolates.
The fourth student picks 2n chocolates.
And so on...
To find the sum of k terms in the above series, we can use the formula for the sum of an arithmetic series: S = (k/2) * (first term + last term)
In this case, the first term is n/2 and the last term is 2^(k/2) * n.
Therefore, the sum of the first k terms is:
S = (k/2) * (n/2 + 2^(k/2) * n)
We need to find the value of k such that the sum of the first k terms is 512.
Solution:
Let's substitute the given values and solve the equation:
512 = (k/2) * (n/2 + 2^(k/2) * n)
Simplifying the equation:
512 = (kn/4) + (2^(k/2) * kn/2)
Multiplying the equation by 4:
2048 = kn + (2^(k/2) * 2kn)
Dividing both sides of the equation by kn:
2048/kn = 1 + 2^(k/2)
Let's check the options given in the question and substitute the values to find the correct option.
Few students are standing one behind the other. They are asked to take...
4 4| 8 16 | 32 64| 128 256
solve by options, for n= 8 first kid takes 4 chocolates.
for even places add all the previous numbers.
for odd 2x the number just before it.
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