Kindly seeking help

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cjiawei

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How do I do the question below?
Given the geometric sequence such that the sum of the first 3 terms is 28 and the sum of its first 6 terms is -133/2 .
Find the 1st term and common ratio of the geometric sequence. Answer: r=-3/2 and a=16

I have done multiple attempts but cant seem to get the workings to the answer.
 
Last edited:
cjiawei said:
How do I do the question below?
Given the geometric sequence such that the sum of the first 3 terms is 28 and the sum of its first 6 terms is -133/2 .
Find the 1st term and common ratio of the geometric sequence. Answer: r=-3/2 and a=16

I have done multiple attempts but cant seem to get the workings to the answer.
Click to expand...
Please post at least one of your attempts so that we know where to begin to help you.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 
Sorry I am unsure if I am correct but this is where i got to so far...
-By forming 2 eqns I tried to subsitute a into the other eqn
- From that point i am unsure on how to further simplify the eqn to get the value 'r'
1597398374493.png

Appreciate the help :)
 
cjiawei said:
Sorry I am unsure if I am correct but this is where i got to so far...
-By forming 2 eqns I tried to subsitute a into the other eqn
- From that point i am unsure on how to further simplify the eqn to get the value 'r'
View attachment 21028

Appreciate the help :)
Click to expand...
What do you get if you divide S6 by S3?
 
First, do you know what a "geometric sequence" is? I would like to think so but you say nothing to let us know in your post. A geometric sequence depends on two numbers, the first number in the sequence, a, and a common ratio, r. The sequence is then \(\displaystyle a, ar, ar^2, ar^3, ....\).

The sum of the first three terms is \(\displaystyle a+ ar+ ar^2\) and we are told that is 28. The sum of the first 6 terms is \(\displaystyle a+ ar+ ar^2+ ar^3+ ar^4+ ar^5\) and you are told that is equal to -133/2.

So you want to solve the two equations
\(\displaystyle a+ ar+ ar^2= 28\) and
\(\displaystyle a+ ar+ ar^2+ ar^3+ ar^4+ ar^5= -\frac{133}{2}\)
for a and r.

And notice that \(\displaystyle ar^3+ ar^4+ ar^5= r^3(a+ ar+ ar^2)\)
 
Thank you for your help. I was mistaken as the formula that i was working for for the sum of geometric sequence was a(1-r^n)/1-r. Nevertheless, thanks for your help.
 
cjiawei said:
Thank you for your help. I was mistaken as the formula that i was working for for the sum of geometric sequence was a(1-r^n)/1-r. Nevertheless, thanks for your help.
Click to expand...
The formula you have is fine, and your work is okay as far as you went, even though others have suggested quicker routes to an answer.

To continue, remove the "(a)" in your last equation, which I don't think you meant to be there, and then simplify the left side. Some things will cancel, particularly after you factor 1 - r^6 as a difference of squares, and then you can solve for r.

Once you see how this works, you may be able to see what the suggestions you were given mean.

But to find a, just plug your value for r into one of your initial equations.
 
While, yes, this is a geometric series, you don't really need that formula! As I pointed out above, ar^3+ ar^4+ ar^5= r^3(a+ ar^2+ ar^3) so a+ ar+ ar^2+ ar^3+ ar^4+ ar^5= (a+ ar+ ar^2)+ r^3(a+ ar+ ar^2)= (1+ r^3)(a+ ar+ ar^2)= -133/2 while we were told that a+ ar+ ar^2= 28.

So we can write (1+ r^3)(28)= -133/2. 1+ r^3= -133/56
r^3= -189/56= -27/8=-3^3/2^3. r= -3/2. Putting that into a+ ar+ ar^2= 28, we get a- 3a/2+ 9a/4= 7a/4= 28. a= 49.
 
cjiawei said:
Thank you for your help. I was mistaken as the formula that i was working for for the sum of geometric sequence was a(1-r^n)/1-r. Nevertheless, thanks for your help.
Click to expand...
Did you solve for "r" from the hints given?
 
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