Video: CBSE Class X • Pack 1 • 2018 • Question 28

CBSE Class X • Pack 1 • 2018 • Question 28

07:22

Video Transcript

The sum of four consecutive numbers in an arithmetic progression is 32, and the ratio of the product of the first and the last terms to the product of the middle two terms is seven to 15. Find the numbers.

Any arithmetic progression or arithmetic series has a first term donated by 𝑎 and a common difference of 𝑑. This means that we can list four consecutive terms as 𝑎, 𝑎 plus 𝑑, 𝑎 plus two 𝑑, and 𝑎 plus three 𝑑. We’re told in the question that the sum of these four terms is 32. Therefore, 𝑎 plus 𝑎 plus 𝑑 plus 𝑎 plus two 𝑑 plus 𝑎 plus three 𝑑 must equal 32. Simplifying this equation by collecting like terms gives us four 𝑎 plus six 𝑑 is equal to 32. As all three of these terms are even, we can divide the equation by two. This gives us two 𝑎 plus three 𝑑 is equal to 16. We will call this equation one.

The product of the first and last terms is 𝑎 multiplied by 𝑎 plus three 𝑑. This can be expanded to 𝑎 squared plus three 𝑎𝑑, as 𝑎 multiplied by 𝑎 is 𝑎 squared and 𝑎 multiplied by three 𝑑 is three 𝑎𝑑. The product of the middle two terms is given by 𝑎 plus 𝑑 multiplied by 𝑎 plus two 𝑑. Expanding these two brackets gives us 𝑎 squared plus two 𝑎𝑑 plus 𝑎𝑑 plus two 𝑑 squared. The first terms, 𝑎 multiplied by 𝑎, gives us 𝑎 squared. The outside terms, 𝑎 multiplied by two 𝑑, gives us two 𝑎𝑑. The inside terms give us 𝑎𝑑. And the last terms multiplied together give us two 𝑑 squared. This can be simplified to 𝑎 squared plus three 𝑎𝑑 plus two 𝑑 squared.

We’re told that these two products are in the ratio seven to 15. This means that multiplying the first product, 𝑎 squared plus three 𝑎𝑑, by 15 will be equal to multiplying the second product by seven. 15 multiplied by 𝑎 squared plus three 𝑎𝑑 is equal to seven multiplied by 𝑎 squared plus three 𝑎𝑑 plus two 𝑑 squared. Expanding the brackets on the left-hand side gives us 15𝑎 squared plus 45𝑎𝑑. And expanding the brackets on the right-hand side gives us seven 𝑎 squared plus 21𝑎𝑑 plus 14𝑑 squared.

As this equation is a quadratic, we need to get it equal to zero to be able to solve it by factorizing. We need to subtract seven 𝑎 squared plus 21𝑎𝑑 plus 14𝑑 squared from both sides of the equation. This gives us eight 𝑎 squared plus 24𝑎𝑑 minus 14𝑑 squared, as 15𝑎 squared minus seven 𝑎 squared is eight 𝑎 squared and 45𝑎𝑑 minus 21𝑎𝑑 is equal to 24𝑎𝑑. Both sides of this equation can be divided by two. This leaves us with four 𝑎 squared plus 12𝑎𝑑 minus seven 𝑑 squared.

We now need to factorize this quadratic equation. Two 𝑎 multiplied by two 𝑎 is equal to four 𝑎 squared. And 𝑑 multiplied by seven 𝑑 is seven 𝑑 squared. As the middle term in our quadratic is positive 12𝑎𝑑, the first bracket needs a negative sign and the second bracket needs a positive sign. We can check this by expanding the two brackets and making sure we get four 𝑎 squared plus 12𝑎𝑑 minus seven 𝑑 squared. Two 𝑎 multiplied by two 𝑎 is four 𝑎 squared. Two 𝑎 multiplied by seven 𝑑 is 14𝑎𝑑. Negative 𝑑 multiplied by two 𝑎 is negative two 𝑎𝑑. And finally, negative 𝑑 multiplied by seven 𝑑 is negative seven 𝑑 squared. The middle two terms positive 14𝑎𝑑 minus two 𝑎𝑑 gives us 12𝑎𝑑. Therefore, the factorization is correct.

This gives us two possible answers, either two 𝑎 is equal to 𝑑 or two 𝑎 is equal to negative seven 𝑑. We now need to substitute each of these into equation one. Firstly, we will substitute two 𝑎 is equal to 𝑑. If two 𝑎 is equal to 𝑑, we are left with 𝑑 plus three 𝑑 is equal to 16. This means that four 𝑑 is equal to 16. Dividing both sides by four gives us a value for 𝑑 equal to four, as 16 divided by four equals four. If 𝑑 is equal to four, two 𝑎 must be equal to four. This means that 𝑎 is equal to two. These values give us an arithmetic progression of two, six, 10 and 14. The first term is equal to two. And the common difference is equal to four.

We will now substitute two 𝑎 is equal to negative seven 𝑑 into equation one. This gives us negative seven 𝑑 plus three 𝑑 is equal to 16. Negative seven 𝑑 plus three 𝑑 is negative four 𝑑. So this is equal to 16. Dividing both sides of this equation by negative four gives us 𝑑 is equal to negative four. If 𝑑 is equal to negative four, two 𝑎 is equal to negative seven multiplied by negative four. This means that two 𝑎 is equal to 28. If two 𝑎 is equal to 28, dividing both sides by two gives us a value of 𝑎 of 14.

This means that we could have the same four numbers but in the opposite order: 14, 10, six, and two. The first term this time is 14 and the common difference is negative four. The four consecutive numbers in the arithmetic progression are two, six, 10 and 14.

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