### 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.