Thread: AS Mathematics Pure Core 2 - Sequences and Series Help!

I need helping forming the equation given in the question, any help?

The sum of the first three terms of a geometric progression is 14. If the first term is 2, and the common ratio is r , show that r² + r - 6 = 0

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I tried substituting a = 2 and n = 3 into the geometric formula for the sum of a series, such that 2(r³ - 1) / (r - 1) = 14, but I can't get it to equal anything near the equation given in the question, so I think I'm doing something wrong...

Originally Posted by LarsaSolidor

I need helping forming the equation given in the question, any help?

The sum of the first three terms of a geometric progression is 14. If the first term is 2, and the common ratio is r , show that r² + r - 6 = 0

---

I tried substituting a = 2 and n = 3 into the geometric formula for the sum of a series, such that 2(r³ - 1) / (r - 1) = 14, but I can't get it to equal anything near the equation given in the question, so I think I'm doing something wrong...

solve for

Thanks for your reply; I managed to work it out about 3 minutes after posting this!