# Video: CBSE Class X • Pack 4 • 2015 • Question 8

CBSE Class X • Pack 4 • 2015 • Question 8

04:37

### Video Transcript

Find the first four terms of the arithmetic progression, where 𝑆 𝑛 denotes the sum of its first 𝑛 terms, 𝑆 five plus 𝑠 seven is 167, and 𝑆 10 is 235.

For an arithmetic progression with first term 𝑎 and common difference 𝑑, the sum of its first 𝑛 terms is 𝑛 over two multiplied by two 𝑎 plus 𝑛 minus one multiplied by 𝑑. For its first five terms, that’s five over two multiplied by two 𝑎 plus five minus one 𝑑. Five minus one is four. So the inside of these brackets simplifies to two 𝑎 plus four 𝑑.

We can also divide though by two. And we get five multiplied by one 𝑎 plus two 𝑑. And then, if we multiply each part of this bracket by the number on the outside, we end up with five 𝑎 plus 10𝑑. Let’s repeat this process for the first seven terms. We get seven over two multiplied by two 𝑎 plus seven minus one 𝑑. Seven minus one is six. And we can once again divide through by two.

Finally, when we expand the brackets by multiplying each term inside the bracket by seven on the outside, we get seven 𝑎 plus 21𝑑. We know that the sum of the first five terms plus the sum of the first seven terms is equal to 167. So we can add these two expressions and equate it to 167. Finally, we should simplify by collecting like terms. That gives us 12𝑎 plus 31𝑑 is equal to 167.

Next, let’s substitute 𝑛 equals 10 to form an equation for the sum of the first 10 terms. This time, we get 10 over two multiplied by two 𝑎 plus 10 minus one multiplied by 𝑑. 10 minus one is nine and 10 divided by two is five. Multiplying the brackets out and we get 10𝑎 plus 45𝑑. We also know though that the sum of the first 10 terms is 235. So we can make this expression equal to 235.

Notice that we now have two equations with two unknowns. We can solve these simultaneously to calculate the value of 𝑎 and 𝑑. First, we need to make the coefficient of either 𝑎 or 𝑑 the same in each equation. Since 12 and 10 have a common multiple of 60, we can multiply the whole of the first equation by five and the whole of the second equation by six. That gives us 60𝑎 plus 155𝑑 equals 835 and 60𝑎 plus 270𝑑 equals 1410.

Since the coefficient of 𝑎 is now the same, we can subtract the first equation from the second. In fact, let’s change the order of these equations so it’s easier to see what’s going on. 60𝑎 minus 60𝑎 is zero, 270𝑑 minus 155𝑑 is 115𝑑, and 1410 minus 835 is 575. To solve this equation, we’ll divide both sides by 115. And that gives us that 𝑑 is equal to five.

We can substitute this value back into either of our equations to calculate the value of 𝑎. If we choose the second, that looks like this: 10𝑎 plus 45 multiplied by five equals 235. 45 multiplied by five is 225. So we can begin to solve this equation by subtracting 225 from both sides. That gives us 10𝑎 is equal to 10, which means that 𝑎 is equal to one.

We’ve calculated the first term in our arithmetic progression is one and the difference between each term is five. All we need to do now is calculate the value of the first four terms of this sequence. Now, we could use the formula for the 𝑛th term in the sequence 𝑎 plus 𝑛 minus one 𝑑. However, we know that the first term is one and the difference between each term is five. So we can repeatedly add five to get the subsequent terms.

The first term is one, the second term is six, the third term is 11, and the fourth term is 16.