Question Video: Using the Properties of Arithmetic Series to Find an Arithmetic Sequence | Nagwa Question Video: Using the Properties of Arithmetic Series to Find an Arithmetic Sequence | Nagwa

Question Video: Using the Properties of Arithmetic Series to Find an Arithmetic Sequence Mathematics

Find the 12-term arithmetic sequence given the sum of the first 4 terms is βˆ’158 and the sum of the last 4 terms is βˆ’574.

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Video Transcript

Find the 12-term arithmetic sequence given the sum of the first four terms is negative 158 and the sum of the last four terms is negative 574.

We know that our arithmetic sequence has a total of 12 terms. We are told that the first four terms sum to negative 158. This means that the sum of the first, second, third, and fourth terms is negative 158. The first term of any arithmetic sequence is equal to π‘Ž. As the common difference is equal to 𝑑, the second term will be π‘Ž plus 𝑑. The third term is π‘Ž plus two 𝑑, and the fourth term is π‘Ž plus three 𝑑. This leads us to the formula that the 𝑛th term, or π‘Ž sub 𝑛, is equal to π‘Ž plus 𝑛 minus one multiplied by 𝑑.

Simplifying our equation by collecting like terms, we get four π‘Ž plus six 𝑑 is equal to negative 158. Dividing both sides of this equation by two gives us two π‘Ž plus three 𝑑 is equal to negative 79. We will call this equation one.

We are also told that the sum of the last four terms is negative 574. As there are 12 terms in total, this will be the ninth, 10th, 11th, and 12th terms. π‘Ž plus eight 𝑑 plus π‘Ž plus nine 𝑑 plus π‘Ž plus 10𝑑 plus π‘Ž plus 11𝑑 is equal to negative 574. This simplifies to four π‘Ž plus 38𝑑 is equal to negative 574. Once again, we can divide both sides of this equation by two, giving us two π‘Ž plus 19𝑑 is equal to negative 287. We will call this equation two. And we now have a pair of simultaneous equations that we can solve by elimination.

We can subtract equation one from equation two to eliminate π‘Ž. On the left-hand side, two π‘Ž minus two π‘Ž is zero and 19𝑑 minus three 𝑑 is 16𝑑. Negative 287 minus negative 79 is equal to negative 208. Dividing both sides by 16, we get 𝑑, the common difference, equal to negative 13. We can now substitute this value back in to equation one.

This gives us two π‘Ž plus three multiplied by negative 13 is equal to negative 79. Three multiplied by negative 13 is negative 39. We can add 39 to both sides so that two π‘Ž is equal to negative 40. Dividing by two, we get the first term π‘Ž equal to negative 20. As the first term is equal to negative 20 and the common difference is negative 13, our arithmetic sequence is negative 20, negative 33, negative 46, and so on.

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