Question Video: Finding the Sum of a Finite Geometric Sequence Mathematics

Video Transcript

The number of terms of a geometric sequence whose first term is 729, last term is one, and sum of all terms is 1093 is blank.

We’re told that the first term of our sequence 𝑎 sub one or 𝑎 is 729. The last term 𝑎 sub 𝑛 is equal to one. We know that 𝑎 sub 𝑛 is equal to 𝑎 multiplied by 𝑟 to the power of 𝑛 minus one. We are also told that the sum of all the terms 𝑆 sub 𝑛 is equal to 1093, where 𝑆 sub 𝑛 is equal to 𝑎 multiplied by one minus 𝑟 to the power of 𝑛 all divided by one minus 𝑟. Our aim in this question is to calculate the number of terms 𝑛.

Substituting in the value of 𝑎, we see that 729 multiplied by 𝑟 to the power of 𝑛 minus one is equal to one. Dividing both sides of this equation by 729, 𝑟 to the power of 𝑛 minus one is equal to one over 729. Using one of our laws of exponents or indices, we can rewrite the left-hand side as 𝑟 to the power of 𝑛 over 𝑟 to the power of one, as 𝑥 to the power of 𝑝 minus 𝑞 is equal to 𝑥 to the power of 𝑝 divided by 𝑥 to the power of 𝑞. We can then multiply both sides of this equation by 𝑟, giving us 𝑟 sub 𝑛 is equal to one over 729 𝑟.

We will now clear some space and consider the second formula. This gives us 729 multiplied by one minus one over 729 𝑟 all divided by one minus 𝑟 is equal to 1093. We can distribute the parentheses of the numerator on the left-hand side to give us 729 minus 𝑟. Multiplying through by one minus 𝑟 gives us 729 minus 𝑟 is equal to 1093 multiplied by one minus 𝑟. We can once again distribute the parentheses or expand the brackets giving us 1093 minus 1093𝑟. Subtracting 729 and adding 1093𝑟 to both sides gives us 1092𝑟 is equal to 364. We can then divide both sides by 1092 such that 𝑟 is equal to one-third. We can now substitute this back in to calculate the value of 𝑛.

One-third to the power of 𝑛 is equal to one over 729 multiplied by one-third. We know that three to the power of six is equal to 729. This means that one-third to the power of six is equal to one over 729. This means that we can rewrite the right-hand side of our equation as one-third to the power of six multiplied by one-third. Once again, using our laws of exponents, we can add the powers. Six plus one is equal to seven. As one-third to the power of 𝑛 is equal to one-third to the power of seven, then 𝑛 must be equal to seven. The number of terms of the geometric sequence whose first term is 729, last term is one, and sum of all terms is 1093 is seven.