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## 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

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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 42 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.