1.1+2.2+3.3+4.4+5.5.............+100.100=?
Before 100.100 we have 99.99?
this is an arithmetic series, where
a=1.2
r=0.1
n = ? but t(n) = 100.100
t(n) = a + (n-1)d
100.100 = 1.1 + .1(n-1)
99=.1n-.1
.1n=99.1
n=991
so there are 991 terms
sum(991) = (991/2)(first + last)
= (991/2)(1.1 + 100.100) = 50144.6
This is not arithmetic series if we have
...9.9+10.10+...+99.99+100.100
correct, I made a silly error,
the d = 1.1 instead of 0.1 the way I had it.
It is an arithmetic series
new solution:
t(n) = 1.1 + (n-1)(1.1)
100.100 = 1.1 + 1.1n - 1.1
n = 91
so there are 91 terms
sum(91) = (91/2)(1.1+100.100) = 4604.6
To find the sum of the given series, you need to add up all the numbers from 1.1 to 100.100.
However, calculating this sum manually can be time-consuming. To simplify the process, you can use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is:
S = (n/2) * (a + l)
Where:
S = the sum of the series
n = the number of terms in the series
a = the first term
l = the last term
In this case, the series is an arithmetic series with a common difference of 1.1. The first term (a) is 1.1, and the last term (l) is 100.100. The number of terms (n) can be found by dividing the difference between the last term and the first term by the common difference and adding 1.
Using these values, we can find the sum of the series:
First, let's find the number of terms:
n = (l - a) / d + 1
= (100.100 - 1.1) / 1.1 + 1
= 100.0 / 1.1 + 1
= 90.9091 + 1
= 91.9091
Now, we can use the formula to find the sum of the series:
S = (n/2) * (a + l)
= (91.9091 / 2) * (1.1 + 100.100)
= 45.95455 * 101.2
= 4640.0726
Therefore, the sum of the given series is 4640.0726.