Video Transcript
Find the arithmetic sequence whose 20th term is 28, given that the sum of its third and sixth terms is greater than its ninth term by eight.
The first term of any arithmetic sequence is denoted by the letter 𝑎. And the common difference is equal to 𝑑. This is the difference between each of the terms. The second term of any arithmetic sequence is therefore equal to 𝑎 plus 𝑑. Adding another 𝑑 to this means that the third term is 𝑎 plus two 𝑑. This pattern continues such that the 𝑛th term is equal to 𝑎 plus 𝑛 minus one multiplied by 𝑑.
We’re told in this question that the 20th term is equal to 28. This means that 𝑎 plus 19𝑑 equals 28. We’re also told that the sum of the third and sixth terms is greater than the ninth term by eight. This means that the third term plus the sixth term is equal to the ninth term plus eight. This can be written as 𝑎 plus two 𝑑 plus 𝑎 plus five 𝑑 is equal to 𝑎 plus eight 𝑑 plus eight. Collecting like terms on the left-hand side gives us two 𝑎 plus seven 𝑑. Subtracting 𝑎 plus eight 𝑑 from both sides of this equation gives us 𝑎 minus 𝑑 is equal to eight, as two 𝑎 minus 𝑎 is 𝑎 and seven 𝑑 minus eight 𝑑 is negative 𝑑.
We now have a pair of simultaneous equations that we can solve to calculate the values of 𝑎 and 𝑑. When we subtract equation two from equation one, the 𝑎’s cancel. Subtracting negative 𝑑 is the same as adding 𝑑. And 19𝑑 plus 𝑑 is equal to 20𝑑. 28 minus eight is equal to 20. Dividing both sides of this equation by 20 gives us a value of 𝑑 equal to one. The common difference of the sequence is one. We can then substitute this value into equation one or equation two to calculate the value of 𝑎. Substituting into equation two gives us 𝑎 minus one is equal to eight. Adding one to both sides of this equation gives us 𝑎 is equal to nine.
As the first term is equal to nine and the common difference is one, our arithmetic sequence is nine, 10, 11, and so on.
