Question

A man ate 100 bananas in five days, each day eating 6 more than the previous day. how many bananas did he eat on the first day?

Answer 1

Suppose he ate x bannanas on the first day then,
A/q
=>x+x+6+x+6+6+x+6+6+6+x+6+6+6+6=100
=>5x+60=100
=>5x=40
=>x=8
Hence that guy ate 8 bannanas on the first day

Answer 2

Given,

A man ate 100 bananas in five days, each day eating 6 more than the previous day.

To find,

The number of bananas he ate on the first day.

Solution,

We can easily solve this problem by following the given steps.

According to the question,

We have the following statements:

A man ate 100 bananas in five days, each day eating 6 more than the previous day.

Let's take the number of bananas on the first day to be x.

So,

x, (x+6), (x+12), ---

This is an AP because the difference between the two consecutive terms is the same.

The first term (a) = x

Common difference(d) = second term - first term

d = x+6-x

d = 6

We know that the following formula is used to find the sum of n terms of an AP:

Sn = n/2 [2a+(n-1)d]

100 = 5/2 [2x+(5-1)6]

5/2 [2x+(4)6] = 100

5/2 (2x+24) = 100

5(x+12) = 100

Dividing both the sides by 5:

(x+12) = 100/5

(x+12) = 20

Moving 12 from the left-hand side to the right-hand side,

x = 20-12

x = 8

Hence, the number of bananas he ate on the first day is 8.