The portal has been deactivated. Please contact your portal admin.

Question Video: Finding the Number of Terms of a Geometric Sequence given Its First and Last Terms and the Sum of All the Terms Mathematics • 10th Grade

Find the number of terms in the geometric sequence given the first term is 21, the last term is 1 5/16, and the sum of all the terms is 40 11/16.

04:56

Video Transcript

Find the number of terms in the geometric sequence given the first term is 21, the last term is one and five sixteenths, and the sum of all the terms is 40 and eleven sixteenths.

In this question, we’re asked to find the number of terms in a geometric sequence. Well, we’re given some properties of the geometric sequence. We’re told its first term is 21, its last term is one and five sixteenths, and the sum of all terms is 40 and eleven sixteenths. To answer this question, we start by recalling a geometric sequence is one where the ratio of successive terms remains constant. And in particular, if the geometric sequence has initial term π‘Ž and ratio of successive terms π‘Ÿ, then we generate the next term in the sequence by multiplying the previous by π‘Ÿ. We get the sequence π‘Ž, π‘Žπ‘Ÿ, π‘Žπ‘Ÿ squared, π‘Žπ‘Ÿ cubed, and so on.

In this question, we’re told the first term of this geometric sequence is 21. So, our value of π‘Ž is 21. And we’re told the last term of this geometric sequence is one and five sixteenths. So, if we say that our geometric sequence has 𝑛 terms, then one and five sixteenths will be π‘Ž times π‘Ÿ to the power of 𝑛 minus one since this is the expression for the 𝑛th term of the sequence. And of course, we know our value of π‘Ž is 21. So, we can rewrite this as 21 times π‘Ÿ to the power of 𝑛 minus one is equal to one and five sixteenths, where we’re going to rewrite this as an improper fraction; it’s 21 divided by 16.

We might be tempted to try and solve this equation for π‘Ÿ. However, this is an equation in two unknowns; we don’t know the value of π‘Ÿ and we don’t know the value of 𝑛. In fact, there’s an infinite number of solutions, so we need more information to determine these values. We need to use the third piece of information we’re given in the question. The sum of all of the terms of this geometric sequence is 40 and eleven sixteenths. And we can do this by recalling the following formula for the sum of the first 𝑛 terms of a geometric sequence. This is given by 𝑆 sub 𝑛 is equal to π‘Ž times one minus π‘Ÿ to the 𝑛th of power all divided by one minus π‘Ÿ provided the ratio π‘Ÿ is not equal to one. We want to apply this to find an expression for the sum of the first 𝑛 terms of this geometric sequence.

However, to do this, we do need to check if π‘Ÿ is equal to one. And we can see that π‘Ÿ is not equal to one. Because if π‘Ÿ was equal to one, then every single term in our geometric sequence would be equal to π‘Ž. However, we’re told in the question the first term is 21, but the last term is one and five sixteenths. These are not equal, so π‘Ÿ cannot be equal to one. Therefore, we can apply this formula to find the sum of the first 𝑛 terms. This gives us that 40 and eleven sixteenths is equal to 21 times one minus π‘Ÿ to the 𝑛th of power all divided by one minus π‘Ÿ. And as we did before, we’re going to write this as an improper fraction. It’s 651 divided by 16.

We now have two equations in two variables, so we need to solve these for the values of π‘Ÿ and 𝑛. There’s a few different ways of doing this. We’re going to start by rearranging this equation. We’ll start by dividing through by 21. This gives us π‘Ÿ to the power of 𝑛 minus one is equal to one divided by 16. Now that we have an expression for π‘Ÿ to the power of 𝑛 minus one, we want to substitute this into our other equation. And we can do this by noticing π‘Ÿ to the 𝑛th power is π‘Ÿ multiplied by π‘Ÿ to the power of 𝑛 minus one. Therefore, if we substitute π‘Ÿ to the power of 𝑛 minus one is equal to one divided by 16 into this equation, we get 651 divided by 16 is equal to 21 multiplied by one minus π‘Ÿ times one over 16 all divided by one minus π‘Ÿ.

Now, we have an equation entirely in terms of π‘Ÿ. So, we can solve this equation for π‘Ÿ. First, we’re going to divide through by 21. This gives us the following. Next, we can cross multiply or multiply both sides of the equation through by one minus π‘Ÿ and both sides of the equation through by 16. This gives us 31 times one minus π‘Ÿ is equal to 16 multiplied by one minus π‘Ÿ over 16. We can then distribute over our parentheses and simplify. We get 31 minus 31π‘Ÿ is equal to 16 minus π‘Ÿ. And now, we can just rearrange and solve for π‘Ÿ. We get 15 is equal to 30π‘Ÿ. And then, we divide through by 30 to see that π‘Ÿ is equal to one-half. And we can substitute this value of π‘Ÿ back into one of our equations to find the value of 𝑛. For example, we can show that one-half raised to the power of 𝑛 minus one must be equal to one over 16.

And there’s many different ways of solving this for 𝑛. One way is to rewrite one over 16 as one-half all raised to the fourth power. Therefore, we have one-half raised to the power of 𝑛 minus one is equal to one-half raised to the fourth power. There’s a few different ways of solving this. We could take log base two of both sides of the equation or log base one-half of both sides of the equation. Or we can use rules of exponents. We know the base one-half is positive and it’s not equal to one. Therefore, for both sides of the equations to be equal, the exponents of both sides of the equation must be equal. 𝑛 minus one must be equal to four, which, of course, proves 𝑛 is five. And of course, 𝑛 is the number of terms of the geometric sequence which we we’re asked to determine. Therefore, the answer to this question is five.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.