# 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 × 𝑛 C₁₀, 4 × 𝑛 C₁₁, 6 × 𝑛 C₁₂ is an arithmetic sequence, find all possible values of 𝑛.

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### Video Transcript

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

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, 𝑎 sub three, and the second term, 𝑎 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 𝑛 choose 12 minus four multiplied by 𝑛 choose 11 is equal to four multiplied by 𝑛 choose 11 minus three multiplied by 𝑛 choose 10. We can rearrange and simplify this equation by adding four multiplied by 𝑛 choose 11 and three multiplied by 𝑛 choose 10 to both sides.

On the left-hand side, we have six multiplied by 𝑛 choose 12 plus three multiplied by 𝑛 choose 10. And on the right-hand side, we have eight multiplied by 𝑛 choose 11. Next, we recall that when dealing with combinations, 𝑛 choose 𝑟 is equal to 𝑛 factorial over 𝑟 factorial multiplied by 𝑛 minus 𝑟 factorial. 𝑛 choose 12 is therefore equal to 𝑛 factorial over 12 factorial multiplied by 𝑛 minus 12 factorial. And multiplying this by six, our first term becomes six multiplied by 𝑛 factorial over 12 factorial multiplied by 𝑛 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 𝑛 factorial. So we can divide through by this. We can simplify this further by recalling that 𝑛 factorial can be written as 𝑛 multiplied by 𝑛 minus one factorial. And this is also equal to 𝑛 multiplied by 𝑛 minus one multiplied by 𝑛 minus two factorial, where 𝑛 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 𝑛 minus 12 factorial. In the same way, the second term can be rewritten as three over 10 factorial multiplied by 𝑛 minus 10 multiplied by 𝑛 minus 11 multiplied by 𝑛 minus 12 factorial. And the sum of these two terms is equal to eight over 11 multiplied by 10 factorial multiplied by 𝑛 minus 11 multiplied by 𝑛 minus 12 factorial.

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

We are now in a position to eliminate the denominators by multiplying through by 22. Our equation becomes 𝑛 squared minus 21𝑛 plus 110 plus 66 is equal to 16𝑛 minus 160. And by subtracting 16𝑛 and adding 160 to both sides, we have the quadratic equation 𝑛 squared minus 37𝑛 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 𝑛 minus 21 multiplied by 𝑛 minus 16 is equal to zero. And as at least one of the expressions in the parentheses must equal zero, we have two solutions 𝑛 equals 21 and 𝑛 equals 16. If the three given terms form an arithmetic sequence, then the two possible values of 𝑛 are 21 and 16. We can check these by substituting them back in to the sequence.