In round 1 of a phone tres a person calls four people n round two ea person calls 4 more people and so on how nany calls will be nade by roynd 5

Question

Help quick

Answer 1

Hi Mikk,

 

I am not sure if I understood the question correctly. 

 

First interpretation of question:

in Round 1, the person called 4 people and in the next rounds, the 4 people will call another 4 people each, and so on. 

 

If that is the case, number of calls in:

round 5 is 4⁵

round 4 is 4⁴

round 3 is 4³

round 2 is 4^{2 and}

round 1 is 4.

 

Note that this is a geometric progression. So when we find the sum, we are finding the geometric series. The formula to find geometric series up to n terms is:

(a(1-rⁿ))/(1-r),

where a is the first term and r is the common ratio of the sequence.

 

So when you add them up:

4+4²+4³+4⁴+4⁵ .

first term, a = 4,

common ratio = 4

number of terms = 5

 

Applying the formula:

4+4²+4³+4⁴+4⁵

= (4(1-4⁵))/(1-4)

=1364

 

The second interpretation of the question could be one person called 4 people in round 1 and in round 2 he called 4 more people, which means he called 8 people in round 2, and in round, he called 4 more people than the previous round.

 

If this is the case, number of calls in

round 1: 4

round 2: 8 i.e. 2*4

round 3: 3*4

round 4: 4*4

round 5: 5*4 = 20

In this case, this is an arithmetic progression. So if we add them up, it will be an arithmetic series. To add them up to n terms, the formula is 

(n/2)(a+l) or (n/2)(2a+(n-1)d)

where a is the first term, l is the last term of the series and d is the common difference.

 

For this question:

total number of calls by round 5

= (5/2)(4+20)

= 60

the other formula..

total number of calls by round 5

=(5/2)(2*4 + (5-1)4)

=60

 

Let me know if i interpreted the question correctly.