Video: CBSE Class X • Pack 2 • 2017 • Question 23

CBSE Class X • Pack 2 • 2017 • Question 23

06:51

Video Transcript

If the ratio of the sum of the first 𝑛 terms of two arithmetic progressions is seven 𝑛 plus one to four 𝑛 plus 27, find the ratio of their ninth terms.

Let’s start by looking at what an arithmetic progression is: a sequence of numbers in which each number differs from the previous one by a constant quantity. And this constant quantity is called the common difference.

Now every arithmetic progression will have a first term, which we’ll call 𝑎, and a common difference, which we’ll call 𝑑. So in this arithmetic progression, the first term is 𝑎. The second term in this progression will differ from the first term by the common difference, which is 𝑑, meaning that the second term will be 𝑎 plus 𝑑. The next term again will differ by 𝑑, and so it will be 𝑎 plus 𝑑 plus 𝑑, which is simply 𝑎 plus two 𝑑. The next term again differs by 𝑑 and so is 𝑎 plus three 𝑑. And we could keep adding 𝑑 to each subsequent term in order to find any term that we wanted to in this progression.

There is in fact a formula for finding the 𝑛th term of an arithmetic progression. If we let the 𝑛th term be equal to 𝑇, then 𝑇 is equal to 𝑎 plus 𝑛 minus one 𝑑. So, for example, if we wanted to find a fifth term of this progression, we simply substitute 𝑛 equals five into this equation. And this gives 𝑎 plus five minus one 𝑑, which can be simplified to 𝑎 plus four 𝑑.

Now if we number the terms which we found earlier, we can see that this is exactly the same as the fifth term in the progression which we found earlier. There’s also an equation for the sum of the first 𝑛 terms of an arithmetic progression. If we let the sum of the first 𝑛 terms be equal to 𝑆, then 𝑆 is equal to 𝑛 over two timesed by two 𝑎 plus 𝑛 minus one times 𝑑.

Now in the question, we’re told that we have two arithmetic progressions. However, we do not know the first term or the common difference of either. So for the first arithmetic progression, we can let the first term be 𝑎 one and we can let the common difference be 𝑑 one. And for the second arithmetic progression, we can let the first term be 𝑎 two and the common difference be 𝑑 two.

Now we can write down formulas for the 𝑛th term of each of these progressions. For the first progression, if we let the 𝑛th term be equal to 𝑇 one, then 𝑇 one is equal to 𝑎 one plus 𝑛 minus one 𝑑 one. And in the second progression, if we let the 𝑛th term be equal to 𝑇 two, then 𝑇 two is equal to 𝑎 two plus 𝑛 minus one 𝑑 two.

And now we can find the ninth term of each of these progressions by substituting 𝑛 equals nine into both of these formulas. And now we can say that the ratio of the ninth term of the first progression to the ninth term of the second progression is 𝑎 one plus eight 𝑑 one to 𝑎 two plus eight 𝑑 two.

Next, we’ll be using the formula for the sum of the first 𝑛 terms in order to find the ratio of the sum of the first 𝑛 terms of these two arithmetic progressions. So we have that the sum of the first 𝑛 terms of the first arithmetic progression, 𝑆 one, is equal to 𝑛 over two times two 𝑎 one plus 𝑛 minus one times 𝑑 one. And the sum of the first 𝑛 terms of the second arithmetic progression, 𝑆 two, is equal to 𝑛 over two timesed by two 𝑎 two plus 𝑛 minus one times 𝑑 two. And so we can say that the ratio of the sum of the first 𝑛 terms of these two arithmetic progressions is 𝑛 over two times two 𝑎 one plus 𝑛 minus one 𝑑 one to 𝑛 over two times two 𝑎 two plus 𝑛 minus one 𝑑 two.

And now we can simplify this ratio by dividing both sides by 𝑛. And since we’re dividing both sides of the ratio by the same amount, the ratio will remain equivalent. Now we can multiply the brackets on the left and the brackets on the right by one-half. This leaves us with the ratio 𝑎 one plus 𝑛 minus one over two 𝑑 one to 𝑎 two plus 𝑛 minus one over two 𝑑 two. And now this ratio which we’ve just found is the ratio of the sum of the first 𝑛 terms of these two arithmetic progressions.

However, we are also given this ratio in the question. It’s given as seven 𝑛 plus one to four 𝑛 plus 27. And so this ratio given in the question will be equivalent to the ratio which we’ve just found. And so we can say that these two ratios are equivalent.

I’m just going to clear up the screen a little bit to give us a bit more space to work with. Next, we’ll be using the ratio of the ninth terms of the progressions in order to find a value of 𝑛 such that- such that this ratio of the ninth terms is equal to the ratio of the sum of the first 𝑛 terms of the two arithmetic progressions.

So we’re trying to make these two ratios equivalent. And we can quite clearly see that these two ratios become equivalent when 𝑛 minus one over two is equal to eight. So now we can solve this equation for 𝑛 and multiply both sides by two and then add one to both sides to give us a value of 𝑛 of 17.

And now if we substitute this value of 𝑛 equals 17 into the ratio of the sum of the first 𝑛 terms of the two arithmetic progressions, then it will be equivalent to the ratio of the ninth terms of the two progressions. But since this ratio of the sum of the first 𝑛 terms is equivalent to the ratio given in the question, we can simply substitute 𝑛 equals 17 straight into this ratio. And this gives us the ratio of seven times 17 plus one to four times 17 plus 27. And since we have substituted in this value of 𝑛 equals 17, this ratio is also equivalent to the ratio of the ninth terms of the progressions, which is exactly what the question is asking us to find.

So now the only step that remains is to simplify this ratio. Multiplying seven by 17, we get 119, and so the left of the ratio becomes 119 plus one. And since four times 17 is 68, the right-hand side of the ratio becomes 68 plus 27. And adding the numbers together gives us 120 to 95.

Now the final step in simplifying this ratio is to divide both sides by five. And now we have found the ratio in its most simple form. And this gives us that the ratio of the ninth terms of these two arithmetic progressions is 24 to 19.

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