# Question Video: Finding the Arithmetic Sequence under a Given Condition Mathematics • 9th Grade

Find the arithmetic sequence given πβπββ = 1708, πβπββ β πβπββ = 336 and all terms are positive.

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### Video Transcript

Find an arithmetic sequence given that π sub one times π sub 12 equals 1708, π sub six times π sub 31 minus π sub seven times π sub 26 equals 336 and all terms are positive.

When dealing with an arithmetic sequence, the subscripts tell us the term number. So, π sub one is the first term of this sequence. π sub 12 would be the twelfth term. To solve for an arithmetic sequence, we need two key things, π sub one, the first term, and π, the common difference. Every term can be written in relationship to the first term and the common difference. π sub π, where π is the term number, is equal to the first term plus π minus one times π.

Because weβre not given what any term is equal to, weβre just given these two equations, we want to start by writing everything in relationship to the first term and the common difference. We know that the sixth term times the 31st term minus the seventh term times the 26th term equals 336. But we want to rewrite the sixth term as the first term plus five times the common difference. We can rewrite the 31st term as the first term plus 30 times the common difference. Weβll write the seventh term as π sub one plus six π and the 26th term as π sub one plus 25π. And we know that all of this is equal to 336.

At this point, weβre gonna need to do some expanding. Weβll multiply π sub one times π sub one, which is π sub one squared. And then, π sub one times 30π is 30π sub one π. Five π times π sub one is five π sub one π. Five d times 30π is 150π squared. And weβll be subtracting whatever we expand from the next two terms. π sub one times π sub one is π sub one squared. π sub one times 25π is 25π sub one π. six π times π sub one is six π sub one π. And six π times 25π is 150π squared, all equal to 336.

We can combine like terms, 30π sub one π plus five π sub one π is 35π sub one π. And 25π sub one π plus six π sub one π is 31π sub one π. Now, remember that weβre subtracting, and we need to distribute this subtraction to the three terms in the parentheses. So, weβll have minus π sub one squared minus 31π sub one π minus 150π squared. And we see that some of these terms start dropping out. π sub one squared minus π sub one squared goes away. And positive 150π squared minus 150π squared cancels out as well.

From there, we can subtract 31π sub one π from 35π sub one π, which gives us four π sub one π equals 336. So, we divide both sides by four. And we get π sub one π equals 84. At this point, you might be wondering what on Earth this has to do with finding the sequence. Remember, these terms stand for something. π sub one is the first term, and π is the common difference. So, from this equation, we have reduced it to say that the common difference times the first term equals 84.

Unfortunately, this is not enough information for us to find the sequence. But fortunately we have one more equation. We also know that π sub one times π sub 12 equals 1708. We want to follow the same procedure here. We want to rewrite the twelfth term in terms of the first term and the common difference. The twelfth term can be rewritten as the first term plus 11 times the common difference. Weβll need to multiply π sub one times both of the terms in the parentheses. π sub one squared plus 11π sub one π equals 1708.

And we know what π sub one times π equals. It equals 84. So, we plug in 84. 11 times 84 equals 924. And so, we subtract 924 from both sides of the equation. And we find out that π sub one squared, the first term squared, equals 784. So, we take the square root of both sides. And the square root of 784 is plus or minus 28. But our question specified that all terms are positive, which means that for us, the first term π sub one is positive 28.

Remember, at the beginning, to find the sequence, we need the first term and the common difference. We donβt have the common difference yet. But we remember that the first term multiplied by the common difference is 84. And now that we know that the first term is 28, we can say that 28 times the common difference has to equal 84. So, we divide both sides of the equation by 28. The common difference is equal to 84 divided by 28, which equals three.

And we found the two pieces of information we need to find this sequence. The first term is 28. And to get to the next term, you need to add π, add the common difference. We know that thatβs three. 28 plus three is 31. 31 plus three is 34. And so, we say that the sequence with these conditions is 28, 31, 34 continuing.