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March 27, 2015

March 27, 2015

Posted by **grandhi** on Sunday, August 21, 2011 at 6:39am.

- arithmetic -
**Anonymous**, Sunday, August 21, 2011 at 8:23amBefore 100.100 we have 99.99?

- arithmetic -
**Reiny**, Sunday, August 21, 2011 at 9:19amthis 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

- arithmetic -
**Anonymous**, Sunday, August 21, 2011 at 10:34amThis is not arithmetic series if we have

...9.9+10.10+...+99.99+100.100

- CHECK - arithmetic -
**Reiny**, Sunday, August 21, 2011 at 12:41pmcorrect, 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

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