G.P
- Thread starter anakbulan
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Please share your work/thoughts about this assignment.
If I were to do this problem:
I would define the first term to be "a" and the common ratio to be "r".
Then
T1 = a ; T2 = ar; T3 = ar2....... continue....
This is how I calculated but i didn’t get the answer so....
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To the extent that I can read that image without breaking my neck, you went
[MATH]ar + ar^2 = 12 \ \& \ ar^2 = ar^3 = 6 \implies[/MATH]
[MATH]\dfrac{ar + ar^2}{ar^2 + ar^3} = \dfrac{12}{6}.[/MATH]
Excellent. Then you completely screwed up simplifying the fractions.
[MATH]\dfrac{ar + ar^2}{ar^2 + ar^3} = \dfrac{12}{6} \implies \dfrac{ar(1 + r)}{ar^2(1 + r)} = 2 \implies \dfrac{\cancel a \cancel r \cancel {(1 + r)}}{\cancel a r^{\cancel 2} \cancel {(1 + r)}} = 2 \implies \dfrac{1}{r} = 2 \implies[/MATH]
[MATH]r = WHAT?[/MATH]
[MATH]\therefore a = WHAT?[/MATH]
How do you check your answer?
[MATH]ar + ar^2 = 12 \ \& \ ar^2 = ar^3 = 6 \implies[/MATH]
[MATH]\dfrac{ar + ar^2}{ar^2 + ar^3} = \dfrac{12}{6}.[/MATH]
Excellent. Then you completely screwed up simplifying the fractions.
[MATH]\dfrac{ar + ar^2}{ar^2 + ar^3} = \dfrac{12}{6} \implies \dfrac{ar(1 + r)}{ar^2(1 + r)} = 2 \implies \dfrac{\cancel a \cancel r \cancel {(1 + r)}}{\cancel a r^{\cancel 2} \cancel {(1 + r)}} = 2 \implies \dfrac{1}{r} = 2 \implies[/MATH]
[MATH]r = WHAT?[/MATH]
[MATH]\therefore a = WHAT?[/MATH]
How do you check your answer?
Last edited:
Yeap the r is 1/2. Got the answer already thanks!