# Question Video: Finding the Terms of a Geometric Sequence and the Sum to Infinity under a Given Condition Mathematics • 10th Grade

The 𝑛th term of a geometric sequence is 𝑎_(𝑛), first term is 𝑎₁, and the sum of the first 𝑛 terms is 𝑆_(𝑛). Find the first three terms and the sum to infinity (𝑆_(∞)) of the infinite geometric sequence with 𝑎₂ − 𝑎₆ = 33 and 𝑆₄ = 132.

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### Video Transcript

The 𝑛th term of a geometric sequence is 𝑎 sub 𝑛, first term is 𝑎 sub one, and the sum of the first 𝑛 terms is 𝑆 sub 𝑛. Find the first three terms and the sum to ∞, 𝑆 sub ∞, of the infinite geometric sequence with 𝑎 sub two minus 𝑎 sub six equals 33 and 𝑆 sub four equals 132.

A geometric sequence is one in which the ratio between consecutive terms is constant. In other words, if we know any term in the sequence, we always get to the next term in the same way by multiplying by the same value, which we call the common ratio and denote using the letter 𝑟. We’ve been given various pieces of information about this sequence. The difference between the second and sixth terms is 33, and the sum of the first four terms is 132. Let’s recall some of the key formulae that we’ll need in this question.

First, the general term of a geometric sequence, 𝑎 sub 𝑛, is given by 𝑎 sub one multiplied by 𝑟 to the power of 𝑛 minus one. We take the first term and multiply it by the common ratio 𝑟 𝑛 minus one times. Secondly, the sum of the first 𝑛 terms of a geometric sequence is given by 𝑆 sub 𝑛 is equal to 𝑎 sub one multiplied by one minus 𝑟 to the power of 𝑛 over one minus 𝑟. And finally, if the absolute value of the common ratio 𝑟 is less than one, then the infinite geometric sequence is convergent. And we can find the sum to ∞ using the formula 𝑆 sub ∞ is equal to 𝑎 sub one over one minus 𝑟. Let’s use these formulae and the information given in the question to form some equations.

For the second term, the value of 𝑛 is two. So the second term 𝑎 sub two is equal to 𝑎 sub one multiplied by 𝑟 to the power of two minus one, which is simply 𝑟 to the first power, or 𝑟. For the sixth term, the value of 𝑛 is six. So we have 𝑎 sub six is equal to 𝑎 sub one multiplied by 𝑟 to the power of five. We’re told that 𝑎 sub two minus 𝑎 sub six is equal to 33. So we have the equation 𝑎 sub one 𝑟 minus 𝑎 sub one 𝑟 to the fifth power is equal to 33. We can factor this equation on the left-hand side by 𝑎 sub one 𝑟 to give 𝑎 sub one 𝑟 multiplied by one minus 𝑟 to the fourth power is equal to 33. And this is our first equation.

Next, we’re told that the sum of the first four terms 𝑆 sub four is equal to 132. So substituting 𝑛 equals four into the formula for the sum of the first 𝑛 terms of a geometric sequence gives 𝑎 sub one multiplied by one minus 𝑟 to the fourth power over one minus 𝑟 is equal to 132. This is our second equation. Now, looking at the two equations we’ve written, we note that both equations contain factors of 𝑎 sub one multiplied by one minus 𝑟 to the fourth power. We can rearrange the first equation to give 𝑎 sub one multiplied by one minus 𝑟 to the fourth power is equal to 33 over 𝑟 and rearrange the second to give 𝑎 sub one multiplied by one minus 𝑟 to the fourth power is equal to 132 multiplied by one minus 𝑟.

As the left-hand sides of these two equations are equal, we can then equate the right-hand sides, which will give an equation in 𝑟 only. 33 over 𝑟 is equal to 132 multiplied by one minus 𝑟. We now look to solve this equation for 𝑟. The first step is to multiply both sides by 𝑟, giving 33 equals 132𝑟 multiplied by one minus 𝑟. Next, we distribute the parentheses on the right-hand side, giving 33 is equal to 132𝑟 minus 132𝑟 squared. And then we can collect all the terms on the left-hand side of the equation by adding 132𝑟 squared and subtracting 132𝑟 from both sides to give 132𝑟 squared minus 132𝑟 plus 33 is equal to zero.

We now see that we have a quadratic equation in 𝑟. In fact this equation can be simplified quite significantly because all three coefficients are multiples of 33. Dividing through by 33 then gives the simplified quadratic four 𝑟 squared minus four 𝑟 plus one is equal to zero. Now, in fact this quadratic equation can be solved by factoring. And it is a perfect square. Four 𝑟 squared minus four 𝑟 plus one is equal to two 𝑟 minus one multiplied by two 𝑟 minus one, or simply two 𝑟 minus one squared. We then solve for 𝑟 by setting the expression inside the parentheses equal to zero because if its square is zero, it must itself be zero. So we have two 𝑟 minus one is equal to zero. And then adding one to both sides of this equation and dividing by two gives 𝑟 is equal to one-half.

So we found the value of 𝑟, the common ratio of this geometric sequence. And we should be encouraged at this point because we’re later asked to find the sum to ∞ of this geometric sequence. And in order to be able to do this, we need the absolute value of 𝑟 to be less than one, which is the case when 𝑟 is equal to one-half. Next, we need to find the value of 𝑎 sub one, the first term in the sequence, which we can do by substituting 𝑟 equals one-half into equation one. Doing so gives 𝑎 sub one multiplied by a half multiplied by one minus a half to the fourth power is equal to 33. A half to the fourth power is one over 16, and then one minus one over 16 is 15 over 16. We then multiply 15 over 16 by one-half, which gives 15 over 32. So we have 𝑎 sub one multiplied by 15 over 32 is equal to 33.

To find the value of 𝑎 sub one, we need to multiply both sides of this equation by the reciprocal of 15 over 32, which is 32 over 15. And we have that 𝑎 sub one is equal to 33 multiplied by 32 over 15, which simplifies to 352 over five. So we found the first term in this sequence. We also need to find the next two terms and then the sum to ∞. So let’s create some space to do this. Remember, to get from one term to the next in a geometric sequence, we multiply by the common ratio. So if the first term is 352 over five, the second term is 352 over five multiplied by one-half, which is 176 over five. The next term, the third term, is 176 over five multiplied by one-half, which is 88 over five.

So we found the first three terms, and now we need to calculate the sum to ∞. Remember, the formula for this is 𝑎 sub one over one minus 𝑟, where the absolute value of 𝑟 must be strictly less than one. So we have 352 over five multiplied by one over one minus a half. Now one minus a half is a half, and one divided by a half is two. So this simplifies to 352 over five multiplied by two, which is 704 over five. So we’ve completed the problem. By first calculating the common ratio and the first term in the sequence, we found that the first three terms are 352 over five, 176 over five, and 88 over five. And the sum to ∞ of this infinite geometric sequence is 704 over five.