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30 Jun 2014, 11:13
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f an infinite GP is summable (|r|<1) then the sum is \frac{b_1}{1-r}
can someone please explain what these means with numbers?
i guess what i'm asking for here is a question where we would use this concept
can someone please explain what these means with numbers?
i guess what i'm asking for here is a question where we would use this concept
30 Jun 2014, 11:53
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sagnik2422
The sum of infinite geometric progression with common ratio \(|r|<1\), is \(sum=\frac{b}{1-r}\), where \(b\) is the first term.
For example, the sum of infinite geometric progression 1/2, 1/6, 1/18, 1/54, ... (first term = 1/2, common ratio = 1/3) is \(sum=\frac{b}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{3}}=\frac{3}{4}\).
Questions to practice:
for-every-integer-m-from-1-to-100-inclusive-the-mth-term-128575.html
for-every-integer-k-from-1-to-10-inclusive-the-kth-term-of-88874.html
a-square-is-drawn-by-joining-the-midpoints-of-the-sides-of-a-102880.html
m17q5-72268.html
ax-y-is-an-operation-that-adds-1-to-y-135277.html
Hope it helps.
09 Aug 2014, 20:24
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"Summation
The sum of an infinite GP will be finite if absolute value of r < 1
The general sum of a n term GP with common ratio r is given by \(b_1*\frac{r^n - 1}{r-1}\)
If an infinite GP is summable (|r|<1) then the sum is \(\frac{b_1}{1-r}\)
"
Awesome post! After 4 years, I still want to read your post. Just need to correct the typo I hightlight than your post is perfect. The correct formula is
\(S=b_1\frac{1-r^n}{1-r}\)
not
\(b_1*\frac{r^n - 1}{r-1}\)
The sum of an infinite GP will be finite if absolute value of r < 1
The general sum of a n term GP with common ratio r is given by \(b_1*\frac{r^n - 1}{r-1}\)
If an infinite GP is summable (|r|<1) then the sum is \(\frac{b_1}{1-r}\)
"
Awesome post! After 4 years, I still want to read your post. Just need to correct the typo I hightlight than your post is perfect. The correct formula is
\(S=b_1\frac{1-r^n}{1-r}\)
not
\(b_1*\frac{r^n - 1}{r-1}\)
