A MAN SAVED #8 THE FIRST DAY #16 THE SECOND DAY #32 THIRD DAY AND SO ON.HOW MANY DAYS WOULD IT TAKES HIM TO SAVE A TOTAL SUM OF #2040
8, 16, 32 , ...
looks like GP with a = 8, r = 2
sum(n) = a(r^n - 1)/(r-1)
2040 = 8(2^n - 1)/(2-1)
255 = 2^n - 1
2^n = 256
I know 2^7 = 256
so n = 7
I will take 7 days
2^8 = 256
To solve this problem, we can start by noticing that the amount of money saved each day follows a pattern. It doubles each day, starting from #8.
Let's break down the problem step-by-step to find the number of days it takes to save a total of #2040:
1. On the first day, the man saved #8.
2. On the second day, he saved #16.
3. On the third day, he saved #32.
4. On the fourth day, he saved #64.
5. On the fifth day, he saved #128.
6. On the sixth day, he saved #256.
From the pattern, we can see that the amount of money saved each day is doubling, so the amount saved on the nth day can be calculated using the formula: 2^(n-1).
Now, let's find the value of n for which the total savings sum up to #2040:
Let's assume n is the number of days required. We need to solve the equation:
2^(0) + 2^(1) + 2^(2) + 2^(3) + ... + 2^(n-1) = 2040
Using the formula for the sum of a geometric series:
(1 - r^n)/(1 - r) = 2040,
where r is the common ratio (in this case, 2).
Simplifying the equation:
(2^n - 1)/(2 - 1) = 2040,
2^n - 1 = 2040,
2^n = 2041.
Taking the logarithm base 2 of both sides:
n log 2 = log 2041,
n = log 2041 / log 2,
n ≈ 10.966.
The number of days required to save a total of #2040 is approximately 10.966 days.
However, since we can't have a decimal number of days, we need to round up to the nearest whole number.
Therefore, it will take him 11 days to save a total of #2040.
To find out how many days it would take for the man to save a total sum of #2040, we need to figure out the pattern in which he saves money. From the information given, we can observe that the man saves a progressively increasing amount each day.
On the first day, he saves #8.
On the second day, he saves #16.
On the third day, he saves #32.
To find the pattern, we notice that each day's savings are doubling from the previous day. This indicates that the man saves an amount given by the formula: 2^(n-1), where n is the day number.
To calculate the number of days required to save a total of #2040, we need to find the day number (n) when the total sum of savings exceeds or equals #2040.
Let's solve for n in the following equation:
2^(0) + 2^(1) + 2^(2) + 2^(3) + ... + 2^(n-1) ≥ 2040
Using the formula for the sum of a geometric series, which is given by:
Sum = (first term * (1 - common ratio^n)) / (1 - common ratio)
In our case, the first term is 1 (2^0) and the common ratio is 2. The sum is given as #2040.
Using this information, we can solve for n:
(1 * (1 - 2^n)) / (1 - 2) ≥ 2040
(1 - 2^n) / -1 ≥ 2040
1 - 2^n ≥ -2040
-2^n ≥ -2041
2^n ≤ 2041
Now, we need to find the smallest integer value of n that satisfies the inequality 2^n ≤ 2041. We can do this by trial and error or by using logarithms.
Taking the logarithm base 2 of both sides:
n ≤ log2(2041)
Using a calculator, we find that log2(2041) is approximately 10.975.
Since we're looking for the smallest integer value of n, we round down to 10.
Therefore, it would take 10 days for the man to save a total sum of #2040.