Video: CBSE Class X • Pack 4 • 2015 • Question 11

CBSE Class X • Pack 4 • 2015 • Question 11

04:59

Video Transcript

The 14th term of an arithmetic progression is twice its eighth term. If its sixth term is negative eight, find the sum of its first 20 terms.

Now, there’s a lot of information that’s been given in this question. But first, we need to recall what is meant by an arithmetic progression.

An arithmetic progression is a sequence in which successive terms are found by adding a constant amount. And that amount can be positive or it can be negative. The general term of an arithmetic progression, which we’ll call 𝑇 𝑛, is equal to 𝑎 plus 𝑛 minus one 𝑑, where 𝑎 represents the first term of the sequence and 𝑑 represents the common difference.

So our sequence starts with whatever the value of 𝑎 is. And then to get to the next term, we keep adding 𝑑. So to get to the 𝑛th term, we have to add 𝑛 minus one lots of 𝑑, which is why the general term is 𝑎 plus 𝑛 minus one 𝑑. Now, we’re going to use the information given in the question to form some equations. And first, we’ll use the fact that the sixth term in this arithmetic progression is negative eight.

So if 𝑇 six is equal to negative eight, then we have that negative eight is equal to 𝑎 plus five 𝑑 and that’s just substituting 𝑛 equals six into our general term. So this gives us one equation connecting 𝑎 and 𝑑 and we’ll use the other information in the question to form another.

The other piece of information we’ve been given is that the 14th term is twice the eighth term. So we have 𝑇 14 is equal to two 𝑇 eight. We can substitute 𝑛 equals 14 into our general term to give 𝑇 14, which is 𝑎 plus 13𝑑, and then substitute 𝑛 equals eight into the general term to give 𝑇 eight, which is 𝑎 plus seven 𝑑. So we have the equation 𝑎 plus 13𝑑 is equal to two lots of 𝑎 plus seven 𝑑.

Now, we can simplify this equation. And first, we will expand the bracket on the right so it gives two 𝑎 plus 14𝑑. And then, if we subtract 𝑎 from each side, we’ll have 13𝑑 is equal to 𝑎 plus 14𝑑. And then, finally, if we subtract 13𝑑 from each side, we have zero is equal to 𝑎 plus 𝑑. This is our second equation. And what we have now is a pair of simultaneous equations in 𝑎 and 𝑑, which we can solve in order to find their values.

Now, there are a couple of different methods that we could use to solve these simultaneous equations. But the method I’m going to use is to subtract equation two from equation one. So we have 𝑎 minus 𝑎 which gives us zero, five 𝑑 minus 𝑑 which gives four 𝑑, and negative eight minus zero which gives negative eight. So we have four 𝑑 is equal to negative eight.

We can find the value of 𝑑 by dividing both sides of the equation by four. And it gives 𝑑 is equal to negative two. Now that we know the value of 𝑑, we can substitute into either equation to find the value of 𝑎. But it’s going to be most straightforward to substitute into equation two. So we substitute 𝑑 equals negative two. And it gives 𝑎 minus two is equal to zero, which can be solved easily by adding two to both sides giving 𝑎 is equal to two.

Now that we know the values of 𝑎 and 𝑑 which remember represent the first term and the common difference of the sequence, we can actually answer the question we’ve been asked which is to find the sum of the first 20 terms of this sequence. We have a general formula for finding the sum of the first 𝑛 terms of an arithmetic progression. And it’s 𝑛 multiplied by two 𝑎 plus 𝑛 minus one lots of 𝑑 all divided by two.

So to find the sum of the first 20 terms, we can substitute 𝑛 is equal to 20, 𝑎 is equal to two, and 𝑑 is equal to negative two into this general formula. And it gives that the sum of the first 20 terms is equal to 20 multiplied by four, which is two 𝑎; that’s two times two, plus 19 multiplied by negative two all divided by two.

Now, 19 multiplied by negative two is negative 38. So within the big bracket, we have four minus 38 which is equal to negative 34. And our sum simplifies to 20 multiplied by negative 34 all over two. Now, we’re multiplying by 20 in the numerator, but then dividing by two in the denominator. So we can cancel a factor of two here. And overall, we’re just multiplying by 10.

So we have 10 multiplied by negative 34, which is equal to negative 340. And this gives the sum of the first 20 terms of this arithmetic progression.

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