For an AP, the sum of the first 100 and the sum of the first 10 terms are 10 and 100 respectively. What is the sum of the first 110 terms of the AP.
@wio a easy one
Hmmm, yeah, it's not too hard. Do you need help?
ye just show me the method lol
i keep getting the wrong answer
First, do you know the formula for an AP?
\[ a_n = a_1+(n-1)d \]
\[ S_n=\frac{n}{2}\left(\frac{a_1+a_n}{2}\right) \]
ye i know
Combine to get: \[ \frac{n}{2}\left(\frac{2a_1+(n-1)d}{2}\right) \]
Plugging in \(n\) and \(S_n\) will give you two formulas, which you can use to solve for \(a_1\) and \(d\).
what is your answer
No, I refuse to give you the answer. I have given you the means to solve it.
i cant get the right answer, mine has heaps of fractions
When \(n=100\), \(S_n=10\)
So\[ \frac{100}{2}\left(\frac{2a_1+(100-1)d}{2}\right) = 10 \]
When \(n=10\), \(S_n=100\).
So\[ \frac{10}{2}\left(\frac{2a_1+(10-1)d}{2}\right) = 100 \]
is d -11/25
my answer is -2310
the options are 90, -90, 110, -110, 100
@wio help
the first thing you need to do is to fix wio's formula. it is a little off
Whoops, I divided by two twice.
yea thats why
"what is your answer?" laughing out loud
k if u could help me, that would be greatly appreciated
\[S_n = \frac{n}{2}(2a_1 + (n-1)d)\] \[S_{100} = 10 \implies 50(2a_1 + 99d)=10\tag{1}\] \[S_{10} = 100\implies 5(2a_1+9d) = 100\tag{2}\] two equations and two unknowns, you can solve them
i got 0/..........
@ganeshie8 i still can't get it please help me
can u check that d = -1/5 and a1 = 10.9
the idea is to fist solve \(a_1\) and \(d\), then use thes to find S_100
but mine is still wrong
Ohk you have already solved a and d
but my answer is 0. it is multiple choice and the options are 90, -90, 110, -110, -100
yeah it looks messy, wish there is a better method
did u get 0 as well?
http://www.wolframalpha.com/input/?i=solve+50%282a_1%2B99d%29%3D10%2C+5%282a_1%2B9d%29%3D100
a1 = 10.99 d = -0.22
use them
-110...
i must be really bad but thanks both of u
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