Find the sum of the geometric series three over 22 minus one over 22 plus one over 66 and so on minus one over 1782.
So we’re told that this is a geometric series. Remember, a geometric series is the sum of a given number of terms of a geometric sequence. And the formula we used to find the sum of the first 𝑛 terms of geometric sequence is 𝑎 sub one times one minus 𝑟 to the 𝑛th power all over one minus 𝑟, where 𝑎 sub one is the first term in our sequence and 𝑟 is the common ratio. Well, we can see that 𝑎 sub one here is three over 22. But what’s the common ratio in this sequence? Well, we find the common ratio by dividing any term in the sequence by the term before it. We could, for example, divide negative one over 22 by three over 22. And that gives us negative one-third. And we would’ve, of course, got the same result had we divided one of 66 by negative one over 22.
And so we now know 𝑎 sub one and 𝑟. But we don’t know how many terms are in our sequence. So how do we find the value of 𝑛? Well, we recall the 𝑛th term of geometric sequence is given by 𝑎 one times 𝑟 to the power of 𝑛 minus one. Now here, 𝑎𝑛, the last time in our sequence, is negative one over 1782. So we can say that negative one over 1782 must be equal to three over 22 times negative one over three to the power of 𝑛 minus one for some value of 𝑛. And that’s what we’re looking to find. Let’s divide through by three over 22. So we get negative one over 243 equals negative one-third to the power of 𝑛 minus one.
Now, there are a couple of ways we can solve this. But actually, the easiest is just to spot that three to the fifth power is 243. And so because it’s an odd power, negative three to the fifth power must be negative 243. We can also go further and say that negative one-third to the power of five must be equal to negative one over 243. So in our equation, 𝑛 minus one must be equal to five, which means 𝑛 must be equal to five plus one which is six. And so we quite clearly need to find the sum of the first six terms. It’s 𝑎 sub one, that’s three over 22, times one minus 𝑟 to the 𝑛th power. Well here, that’s negative one-third to the sixth power. And that’s all over one minus 𝑟, one minus negative one-third. And that gives us 91 over 891.
And so we find the sum of the first six terms of the geometric sequence we defined to have a first term of three over 22 and a common ratio of negative one-third and, therefore, the sum of the geometric series in this question to be 91 over 891. Now, there is a way we can check some of the work that we’ve done. We calculated 𝑛 of equal to six. And it might indeed be sensible just to double check that the sixth term in our geometric sequence is indeed negative one over 1782. Well, going back to the formula for the 𝑛th term, we see that the sixth term is three over 22 multiplied by negative one-third to the power of six minus one. Well, that does indeed give us negative one over 1782 as required.