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• Solving sequences an plus#

Solving sequences an plus#

So we have 𝑛 is equal to negative one plus or minus the square root of one squared minus four multiplied by one multiplied by negative 200 all over two multiplied by one. The coefficients in our quadratic equation are one, one, and negative 200. Remember, the quadratic formula tells us that the roots of the equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero are given by 𝑥 equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. This equation can’t be factored, so we’ll solve by applying the quadratic formula.

And finally, we can subtract 200 from each side of the equation to give zero is equal to 𝑛 squared plus 𝑛 minus 200. Distributing the parentheses, on the right-hand side, we have 200 is equal to 𝑛 squared plus 𝑛. We can then multiply both sides of the equation by two to give 200 equals 𝑛 multiplied by 𝑛 plus one. And within the parentheses, this simplifies further to 𝑛 plus one. This equation simplifies to 100 equals 𝑛 over two multiplied by two plus 𝑛 minus one. We have 100 equals 𝑛 over two multiplied by two times one plus 𝑛 minus one multiplied by one. We can therefore form an equation by substituting 100 for 𝑆 sub 𝑛, one for 𝑎, and one for 𝑑. The sum of the first 𝑛 terms we want to be 100 and 𝑛, the number of terms, we don’t know. The first term and the common difference for this sequence are each one. 𝑎 or sometimes 𝑎 sub one represents the first term in the sequence.

𝑛 represents the number of terms whose sum we’re finding. It’s 𝑆 sub 𝑛 equals 𝑛 over two multiplied by two 𝑎 plus 𝑛 minus one 𝑑, where 𝑆 sub 𝑛 represents the sum of the first 𝑛 terms. We recall the formula for finding the sum of the first 𝑛 terms of an arithmetic sequence. That means we’re looking for the day on which the sum of the terms first exceeds 100 pounds. We are asked to find the day on which he will have saved over 100 pounds in total. And so the terms form an arithmetic sequence with a common difference of one. The difference between the terms in this sequence is therefore constant. The amounts Benjamin saves increase by one pounds each day. On the first day, it’s one pound on the second day, it’s two pound on the third day, it’s three pound and so on. Let’s look at the sequence formed by the amounts Benjamin saves each day.

On which day will he have saved over 100 pounds in total? Benjamin saves one pound on the first day, two pound on the second day, three pounds on the third day, and so on, saving an extra one pound each day.