# Question Video: Finding the Value of an Unknown given Three Consecutive Terms of an Arithmetic Series Each in the Form of a Combination Mathematics

Given that 3 Ć š Cāā, 4 Ć š Cāā, 6 Ć š Cāā is an arithmetic sequence, find all possible values of š.

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### Video Transcript

Given that three multiplied by š choose 10, four multiplied by š choose 11, six multiplied by š choose 12 is an arithmetic sequence, find all possible values of š.

We begin by recalling that in any arithmetic sequence, there is a common or constant difference between consecutive terms. This means that the difference between the third term, š sub three, and the second term, š sub two, must be equal to the difference between the second term and the first term. Substituting our three terms, we have six multiplied by š choose 12 minus four multiplied by š choose 11 is equal to four multiplied by š choose 11 minus three multiplied by š choose 10. We can rearrange and simplify this equation by adding four multiplied by š choose 11 and three multiplied by š choose 10 to both sides.

On the left-hand side, we have six multiplied by š choose 12 plus three multiplied by š choose 10. And on the right-hand side, we have eight multiplied by š choose 11. Next, we recall that when dealing with combinations, š choose š is equal to š factorial over š factorial multiplied by š minus š factorial. š choose 12 is therefore equal to š factorial over 12 factorial multiplied by š minus 12 factorial. And multiplying this by six, our first term becomes six multiplied by š factorial over 12 factorial multiplied by š minus 12 factorial.

We can rewrite the other two terms in our equation in the same manner. We notice that each of the three terms has a common factor of š factorial. So we can divide through by this. We can simplify this further by recalling that š factorial can be written as š multiplied by š minus one factorial. And this is also equal to š multiplied by š minus one multiplied by š minus two factorial, where š is greater than or equal to two. Using this property, we can rewrite the denominator of our first term as 12 multiplied by 11 multiplied by 10 factorial multiplied by š minus 12 factorial. In the same way, the second term can be rewritten as three over 10 factorial multiplied by š minus 10 multiplied by š minus 11 multiplied by š minus 12 factorial. And the sum of these two terms is equal to eight over 11 multiplied by 10 factorial multiplied by š minus 11 multiplied by š minus 12 factorial.

All three denominators have a common factor of 10 factorial multiplied by š minus 12 factorial. Clearing some space, our equation simplifies to six over 12 multiplied by 11 plus three over š minus 10 multiplied by š minus 11 is equal to eight over 11 multiplied by š minus 11. The first term has a common factor of six in the numerator and denominator. We can then multiply through by š minus 10 multiplied by š minus 11. After canceling the common factors, we can begin to distribute our parentheses. š minus 10 multiplied by š minus 11 is equal to š squared minus 21š plus 110. And eight multiplied by š minus 10 is equal to eight š minus 80.

We are now in a position to eliminate the denominators by multiplying through by 22. Our equation becomes š squared minus 21š plus 110 plus 66 is equal to 16š minus 160. And by subtracting 16š and adding 160 to both sides, we have the quadratic equation š squared minus 37š plus 336 equals zero. This can be solved using the quadratic formula or by factoring. Since negative 21 and negative 16 sum to negative 37 and the two values have a product of 336, our equation simplifies to š minus 21 multiplied by š minus 16 is equal to zero. And as at least one of the expressions in the parentheses must equal zero, we have two solutions š equals 21 and š equals 16. If the three given terms form an arithmetic sequence, then the two possible values of š are 21 and 16. We can check these by substituting them back in to the sequence.