Exercises 88–90 will help you prepare for the material covered in...

Question

Exercises 88–90 will help you prepare for the material covered in the next section. Consider the sequence whose nth term is an = (3)5^n Find a2/a3, a1/a2, a4/a3 and a5/a4 What do you observe?

Accepted Answer

Hey everyone here we are asked to evaluate the ratios using the equation of the end term of the sequence which is a sub N is equal to five times the quantity of seven to the power of N. So we can begin here Given ratio, we have a sub two divided by a sub one. So we can use our given equation beginning with our numerator. We have a sub two which means we are looking for the second term in the sequence. So our end here is two. So again using our given equation we have five times the quantity of seven to the power of N. Which in this case is just to and then again for our denominator we use this equation, we have five times the quantity of seven. And here we are looking for our first term because we have a sub one. So R N is just one and now we can go ahead and simplify this. So we see that the five in the numerator and denominator cancel out. And so if we recall the exponent rules we can simplify this to seven to the power of two minus one and then seven to the power of two minus one is simply seven to the power of one which is just seven. So now we have our simplified ratio for this a sub two divided by a sub one. So we can move on to the second ratio. Here we have a sub three Divided by a sub two. So here we can again use our given equation. So for our numerator, our end value will be three. So we have five times the quantity of seven to the power of three. And then our denominator is just the same as the numerator from our previous ratio. So we have five times the quantity of seven to the power of two. And now once again we need to simplify this so the fives cancel in both the numerator and the denominator. And now using our exponent rules we have seven to the power of three -2. This just gives us seven to the power of one which is simply seven. Now we can move on to the next ratio where we have a sub four decided by a sub three. So now we just repeat the process is from before. So beginning with our numerator we have five times the quantity of seven to the power here we have our n value of four. So we have to the power of four divided by five times the quantity of seven to the power of three. Now we need to simplify this, our fives cancel and then we are left with seven to the power of 4 -3. 4 -3 is just one. So we are left with the simplified ratio of seven to the power of one which is simply seven. And then we can move on to our last ratio here we have a sub five divided by a sub four. So beginning again with our numerator, we are looking for the fifth term here. So our equation is five times the quantity of seven to the power of five and then divided by our denominator. We have a sub four. So our equation is five times the quantity of seven to the power of four. Once again simplifying our fives cancel in using our exponent rules, we are left with seven to the power of five minus four. Five minus four is just one. So we have seven to the power of one, which leaves us with just seven. So now we can see that we have found each of the ratios here and there are each equal to seven. So the ratio of the term that comes before each ratio is always seven and that leaves us with answer choice. A thanks for watching. I hope you found this video helpful and I'll see you again next time.

Exercises 88–90 will help you prepare for the material covered in the next section. Consider the sequence whose nth term is an = (3)5^n Find a2/a3, a1/a2, a4/a3 and a5/a4 What do you observe?

Key Concepts

Sequences

Ratios

Exponential Growth

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