You are Here: Home >< Maths

1. If the sum of first 100 terms of a GP is 0 and the first term is -1, then the common ratio of GP will be:
i)2
ii)1
iii) -1
iv)1
I generally think that the answer should be 1 but the actual answer is -1. So I tried the whole formula for the sum of the no. of terms of GP and I'm still getting 1. Please help.
2. (Original post by a_09)
If the sum of first 100 terms of a GP is 0 and the first term is -1, then the common ratio of GP will be:
i)2
ii)1
iii) -1
iv)1
I generally think that the answer should be 1 but the actual answer is -1. So I tried the whole formula for the sum of the no. of terms of GP and I'm still getting 1. Please help.
It can’t be 1 otherwise it wouldnt go to 0.

You’d basically just be left with -1 -1 -1 + ... -1 = 0 which is not possible
3. (Original post by RDKGames)
It can’t be 1 otherwise it wouldnt go to 0.

You’d basically just be left with -1 -1 -1 + ... -1 = 0 which is not possible
Can you please show me some working.
4. (Original post by a_09)
If the sum of first 100 terms of a GP is 0 and the first term is -1, then the common ratio of GP will be:
i)2
ii)1
iii) -1
iv)1
I generally think that the answer should be 1 but the actual answer is -1. So I tried the whole formula for the sum of the no. of terms of GP and I'm still getting 1. Please help.
Try summing the first say 6 terms of a sequence with a = -1 and r = -1 and you should notice something. Then compare this with a series with a = -1 and r = 1 (RDKGames has already shown why this won't give you a sum of 0).

If you're using the sum formula then that might produce two answers but remember that you can't divide by 0...

5. (Original post by Notnek)
Try summing the first say 6 terms of a sequence with a = -1 and r = -1 and you should notice something. Then compare this with a series with a = -1 and r = 1 (RDKGames has already shown why this won't give you a sum of 0).

If you're using the sum formula then that might produce two answers but remember that you can't divide by 0...

Here's the working as per the sum formula(I hope I'm not doing it the wrong way) :-
S(100)=(a(r^n)-1)/r-1
0=(-1(r^100)-1)/r-1
0=(r^100)-1
r^100=1
r=1
I get your point, that multiplying it with -1 would result in -1,1,-1,.... and thus the sum will be 0.
6. (Original post by a_09)
Here's the working as per the sum formula(I hope I'm not doing it the wrong way) :-
S(100)=(a(r^n)-1)/r-1
0=(-1(r^100)-1)/r-1
0=(r^100)-1
r^100=1
r=1
I get your point, that multiplying it with -1 would result in -1,1,-1,.... and thus the sum will be 0.
r^100 = 1 has two real solutions. What are they?

r = 1 is invalid as I said in my last post because that would mean that s(100) has a 0 on the denominator which is not possible.

When you solve a fraction equal to 0, you are right to set the numerator equal to 0 but you always should check to see if your solutions are valid.
7. Okay. Yeah, got it. Thank you! 😃

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 7, 2018
Today on TSR

### Edexcel C2 Core Unofficial Markscheme!

Find out how you've done here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams