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Question Video: Finding the Number of Terms of an Arithmetic Sequence given the Sum of All the Terms Under a Given Condition Mathematics • 10th Grade

In an arithmetic sequence that starts at π‘Žβ‚, π‘Žβ‚‚β‚‡ = 29 and the sum of the eighth and the twelfth terms exceeds the fifteenth by 7. Find the number of terms whose sum is 322.

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Video Transcript

In an arithmetic sequence that starts at π‘Ž sub one, π‘Ž sub 27 equals 29 and the sum of the eighth and the 12th terms exceeds the 15th by seven. Find the number of terms whose sum is 322.

Remember, an arithmetic sequence is one which has a common difference between terms. We’re going to work backwards by looking at forming an expression that describes some number of terms of an arithmetic sequence.

Suppose we have an arithmetic sequence with first term π‘Ž and common difference 𝑑. 𝑆 sub 𝑛, which describes the sum of the first 𝑛 terms, is given by 𝑛 divided by two times two π‘Ž plus 𝑛 minus one times 𝑑. Now, in fact, we’re told the first term is π‘Ž sub one. And we want to find the number of terms, the value of 𝑛, where the sum is 322. So let’s let 𝑆 sub 𝑛 be 322. Then, since the first term in our sequence is π‘Ž sub one, the right-hand side becomes 𝑛 divided by two times two π‘Ž sub one plus 𝑛 minus one times 𝑑.

Now, it follows that in order to be able to calculate the value of 𝑛, we’re going to need to work out the value of 𝑑 and the value of π‘Ž sub one. And there’s some information in the question that will allow us to do so. First, we know that the 27th term is 29 and we also know the formula that allows us to find the 𝑛th term in an arithmetic sequence. It’s π‘Ž plus 𝑛 minus one times 𝑑. This time, let’s let π‘Ž sub 𝑛 be π‘Ž sub 27. It’s 29. Then, the right-hand side of this equation can be written as π‘Ž sub one plus 𝑛 minus one β€” that’s 27 minus one β€” 𝑑, or 26𝑑.

The next bit of information compares the sum of the eighth and 12th terms to the 15th. So let’s form expressions for the 15th, eighth, and 12th term. The 15th term is π‘Ž sub one plus 15 minus one 𝑑, or 14𝑑. Then, the eighth and 12th terms are π‘Ž sub one plus seven 𝑑 and π‘Ž sub one plus 11𝑑, respectively. We know that the sum of the eighth and 12th terms exceeds the 15th by seven. So let’s find their sum. It’s π‘Ž sub one plus seven 𝑑 plus π‘Ž sub one plus 11𝑑, which is two π‘Ž sub one plus 18𝑑.

We’re going to relate this expression to the expression for π‘Ž sub 15. We know that the sum of the eighth and 12th terms will be equivalent to the 15th term plus seven. So we can form the following equation. Two π‘Ž sub one plus 18𝑑 equals π‘Ž sub one plus 14𝑑 plus seven. If we subtract π‘Ž sub one and 14𝑑 from both sides, we get the following equation. π‘Ž sub one plus four 𝑑 equals seven.

We might now see that we have a pair of simultaneous equations. We can use this equation along with the equation we formed describing the 27th term. Let’s clear some space to do so.

We can rearrange our second equation so that it matches the form of the first. Then, we notice that the coefficient of π‘Ž sub one is the same. So we can subtract one equation from the other. 29 minus seven is 22, π‘Ž sub one minus π‘Ž sub one is zero, and 26 minus four is 22. So we get 22 equals 22𝑑. Dividing through by 22, and we find 𝑑 is equal to one. And that’s great because we’ll be able to substitute this into the expression we formed to describe the sum of some number of terms.

Before we do though, let’s evaluate π‘Ž sub one. We’re going to substitute this into the equation seven equals π‘Ž sub one plus four 𝑑. When we do, we get seven equals π‘Ž sub one plus four times one, or just π‘Ž sub one plus four. We then subtract four from both sides, and we get π‘Ž sub one equals three.

We now have all the information we need to describe our arithmetic sequence. We know it has a common difference of one and a first term of three. But we also know we want to find the number of terms whose sum is 322. So let’s substitute these values into that equation. That gives us 322 equals 𝑛 over two times two times three plus 𝑛 minus one times one. Evaluating the terms inside our parentheses, and it simplifies simply to five plus 𝑛.

We’re going to write some of the key information down and then clear some space to solve this equation. So, to solve this equation, let’s begin by dealing with the fraction. We’re going to multiply both sides of the equation by two. That gives us 644 equals 𝑛 times five plus 𝑛. We can then distribute the parentheses on the right-hand side, and we get five 𝑛 plus 𝑛 squared.

Here, we notice we have a quadratic equation in 𝑛. And to solve it, we need to set it equal to zero. Let’s subtract 644 from both sides. Then, to factor the expression on the right-hand side, we need to find a pair of numbers whose product is negative 644 and whose sum is five. Perhaps a little bit of trial and error, and we find that we get negative 23 and 28. So this quadratic expression is 𝑛 minus 23 times 𝑛 plus 28. And for the product of these two expressions to be zero, we know that one or other must itself be zero. We then solve each equation for 𝑛, and we get 𝑛 equals 23 or negative 28. And at this stage, we can instantly disregard 𝑛 equals negative 28.

Remember, we’re finding a number of terms, so 𝑛 must be a natural number. This means that 𝑛 is 23. And so it must be 23 terms whose sum is 322. The answer is 23.

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