Video: CBSE Class X • Pack 4 • 2015 • Question 22

CBSE Class X • Pack 4 • 2015 • Question 22

07:06

Video Transcript

Find the 60th term of the arithmetic progression eight, 10, 12, and so on. If it has a total of 60 terms, find the sum of its last 10 terms.

Any arithmetic progression has a first term, 𝑎, and a common difference, 𝑑. We can also use the formula 𝑎 plus 𝑛 minus one multiplied by 𝑑 to work out the 𝑛th term of the sequence or progression. The sequence in this question is eight, 10, 12, and so on. It has a difference of two, as we need to add two to get from eight to 10 and add two to get from 10 to 12.

The first term of the sequence, 𝑎, is equal to eight, and the common difference, 𝑑, is equal to two. Substituting these numbers into the formula can help us calculate the 60th term: eight plus 60 minus one multiplied by two. 60 minus one is equal to 59. Multiplying this by two gives us 118. Finally, adding eight gives us a 60th term of 126. The 60th term in the arithmetic progression eight, 10, 12, and so on is 126.

The second part of our question told us that the progression had a total of 60 terms. We need to find the sum of the last 10 of these terms. The last 10 terms of the sequence will be the 51st, the 52nd, the 53rd, and so on, all the way up to the 60th term. We already know that the 60th term is 126. We are now going to work out the 51st term.

Using the same formula as before, the 51st term is equal to eight plus 51 minus one multiplied by two. 51 minus one is equal to 50, and multiplying this by two gives us 100. This means that the 51st term is 108. As the common difference is two, we could work out the 52nd, 53rd, 54th, and so on terms, all the way up to the 59th. For example, the 52nd term is 110; the 53rd term is 112; the 59th term is 124.

One way to answer the question would be to add these 10 numbers. However, this could be quite time-consuming if we had to add a lot of numbers. An alternative method is to use the formula that the sum of any 𝑛 terms is equal to 𝑛 divided by two multiplied by 𝑎 plus 𝑙.

𝑎 is the first term in the set of numbers we’re looking to sum, in this case 108. 𝑙 is the last term in this set of numbers, in this case 126. And 𝑛 is the number of terms. In this case, we’re looking to sum these 10 terms. This means that the sum of the last 10 terms can be calculated by dividing 10 by two and multiplying this answer by 108 plus 126.

10 divided by two is equal to five, and 108 plus 126 is equal to 234. We now need to multiply 234 by five. Five multiplied by four is equal to 20. We put the zero in the ones or units column and carry the two. Five multiplied by three is 15. Adding the two gives us 17. We put the seven in the tens column and carry the one. Five multiplied by two is equal to 10. Adding the one give us 11. This means that five multiplied by 234 is equal to 1170.

An alternative method to work out the second part of the question would be using the formula 𝑠 of 𝑛 equals 𝑛 divided by two multiplied by two 𝑎 plus 𝑛 minus one multiplied by 𝑑. We could use this formula to work out the sum of the first 60 terms. We could also use it to work out the sum of the first 50 terms. Subtracting these two answers would give us the sum of the last 10 terms, the 51st to the 60th term.

Substituting in our values of 𝑎 and 𝑑, eight and two, we can calculate the sum of the first 60 terms: 60 divided by two multiplied by two multiplied by eight plus 60 minus one multiplied by two. 60 divided by two is equal to 30. Two multiplied by eight is equal to 16. 60 minus one is equal to 59, and multiplying this by two is equal to 118. We’re therefore left with 30 multiplied by 16 plus 118. 16 plus 118 is equal to 134, and multiplying this by 30 gives us 4020. Therefore, the sum of the first 60 terms is equal to 4020.

We can calculate the sum of the first 50 terms in the same way: 50 divided by two multiplied by two multiplied by eight plus 50 minus one multiplied by two. This is equal to 25 multiplied by 16 plus 98. 16 plus 98 is equal to 114, and 114 multiplied by 25 is equal to 2850. This means that the sum of the first 50 terms is 2850. We can calculate the sum of the last 10 terms by subtracting this number from 4020. This is equal to 1170. Therefore, once again, we have proved that the sum of the last 10 terms is 1170.

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