# Video: CBSE Class X • Pack 2 • 2017 • Question 6

CBSE Class X • Pack 2 • 2017 • Question 6

07:08

### Video Transcript

Which term of the progression 20, 19 and a quarter, 18 and a half, 17 and three-quarters, and so on is the first negative term?

We can represent this progression on a number line. The first term is 20, which is here. The second term is 19 and a quarter, which is here. The third term is 18 and a half, which is here. And the fourth term is 17 and three-quarters, which is here.

Can you see the pattern? To get from the first term to the second term, you subtract three-quarters. To get from the second term to the third term, you subtract three-quarters. And to get from the third term to the fourth term, you subtract three-quarters. There is a common difference of negative three-quarters between two consecutive terms of our progression. So this is an arithmetic progression.

We can continue the pattern to find the fifth term of the arithmetic progression. We subtract three-quarters from the fourth term to get to the fifth term, 17.

The question we have to answer is “Which term is the first negative term?” All the terms we’ve seen so far are positive, but they’re getting smaller and smaller. Continuing the pattern then, eventually, we’re going to go past zero on our number line and end up with a negative term. This will be the first negative term and all following terms will also be negative.

The question is “Is this first negative term of our arithmetic progression the 10th term or the 20th term or the 100th term of our arithmetic progression?” That’s what the question is asking us. One way to find this out would be to continue the pattern and find the sixth term and then the seventh term and then the eighth term until we found the term which was negative.

We subtract three-quarters from the fifth term to find the sixth term and we see that it is 16 and a quarter. And we can find the seventh term by subtracting three-quarters from this. But this would take a while. There’s a better way. We use the fact that the 𝑛th term of the arithmetic progression with first term 𝑎 and common difference 𝑑 is 𝑎 subscript 𝑛 equals 𝑎 plus 𝑛 minus one times 𝑑.

What’s the 𝑛th term of our arithmetic progression? Well, the first term 𝑎 is 20. And we found that the common difference 𝑑 was negative three-quarters. Now, we can simplify. We do this by expanding the brackets. We get 𝑛 times negative three-quarters, which is negative three quarters 𝑛. And to this, we add negative one times negative three-quarters. The two minus signs make a plus. And so we get plus three-quarters.

Now, we can write each term as a fraction with denominator four. 20 is 80 over four. Adding negative three-quarters 𝑛 is the same as subtracting three 𝑛 over four. And the plus three over four is already written in the form we need.

Now that all the fractions have the same denominator, it’s simple to combine them. The 𝑛th term of our arithmetic progression is 83 minus three 𝑛 over four. We can check that this formula works. For example, we can find the sixth term of our progression by substituting six for 𝑛. We get 83 minus three times six over four which is 65 over four, which as a mixed number is 16 and a quarter, which is what we found on the number line.

Okay, so we have a formula for the 𝑛th term of our progression. How does this help us find the first negative term? Remember we want to find which is the first negative term. Now, if we knew what the value of this negative term was, for example, if we knew that the first negative term was negative three-quarters or negative one or negative 0.01, we could use this expression for 𝑎 subscript 𝑛 to find the value of 𝑛 and hence which term the first negative term is.

Unfortunately, we don’t know what the first negative term is. What we can do is instead of worrying about which is the first negative term, we can just find when a term is negative. We, therefore, want to solve 𝑎 subscript 𝑛 is less than zero. Solving this inequality for 𝑛 tells us for which 𝑛 the 𝑛th term is negative.

We use our expression for 𝑎 subscript 𝑛, the 𝑛th term. So 83 minus three 𝑛 over four must be less than zero for the 𝑛th term to be negative. We solve this inequality in the normal ways starting by multiplying both sides by four to get that 83 minus three 𝑛 must be less than zero. Adding three 𝑛 to both sides gives us that 83 must be less than three 𝑛. Dividing both sides by three, we find that 83 over three must be less than 𝑛.

We can swap the sides of the inequality, remembering to flip the sign also. 83 over three is less than 𝑛 means that 𝑛 is greater than 83 over three. And finally, as 83 divided by three is 27 remainder two, we can write 83 over three as a mixed number as 27 and two-thirds.

So when is the 𝑛th term negative? The answer is when 𝑛 is greater than 27 and two-thirds. We can write this down explicitly: the 𝑛th term 𝑎 subscript 𝑛 is negative when 𝑛 is greater than 27 and two-thirds. So for example, we can be sure that the 100th term is negative because 100 is certainly greater than 27 and two-thirds.

But we’re interested in the first negative term. So we want the smallest 𝑛. And as 𝑛 is a whole number, the smallest 𝑛 for which 𝑛 is greater than 27 and two-thirds is 𝑛 equals 28. The 28th term is negative because 28 is greater than 27 and two-thirds. But the 27th term and all previous terms are not negative because 𝑛 is not greater than 27 and two-thirds in this case.

You can check this is true by using the formula for the 𝑛th term. Substituting 28 for 𝑛, we find that the 28th term is negative a quarter. And in a similar way, we find that the 27th term is a half. So the 28th term really is the first negative term. The first negative term is the 28th term, which happens to be negative one-quarter.

We can look back to our number line and label this first negative term as the 28th term. The term immediately before it, the 27th term, is positive as are all other terms before the 28th term. And of course, the 29th term and all other terms after the 28th term are negative.