1.1+2.2+3.3+4.4+5.5.............+100.100=?
4 answers
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
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
...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
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