Series, Progressions and the Binomial Expansion: Question 2

Worked solution

Part (a): Number of seats in Row 20

The seat counts form an arithmetic progression with first term $a = 14$ and common difference $d = 3$. The $n$th term is

$u_n = a + (n-1)d.$

For Row $20$ ($n=20$):

$u_{20} = 14 + (20-1)(3) = 14 + 19\times3 = 14 + 57 = 71.$

Check: building the sequence row by row, $14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71$. That is $20$ terms, and the last one is indeed $71$, confirming the formula.

Part (b): Total number of seats in the first 20 rows

$S_n = \frac{n}{2}\bigl(2a + (n-1)d\bigr)$

$S_{20} = \frac{20}{2}\bigl(2(14) + 19(3)\bigr) = 10\times(28+57) = 10\times85 = 850.$

Check: using the equivalent formula $S_n = \frac{n}{2}(u_1+u_n)$ with $u_1=14$ and $u_{20}=71$ (from part (a)):

$S_{20} = \frac{20}{2}(14+71) = 10\times85 = 850.$

Both methods agree, so $S_{20}=850$.

Part (c): Total seating capacity for 30 rows

Now $n=30$. First find $u_{30}$:

$u_{30} = 14 + 29\times3 = 14+87=101.$

Then

$S_{30} = \frac{30}{2}(u_1+u_{30}) = 15\times(14+101)=15\times115=1725.$

Check with the other sum formula:

$S_{30}=\frac{30}{2}\bigl(2(14)+29(3)\bigr)=15\times(28+87)=15\times115=1725.$

Both methods agree, so the theatre’s total seating capacity is $1725$ seats.

Final answers

• (a) Row $20$ has $\boxed{71}$ seats.
• (b) The first $20$ rows contain $\boxed{850}$ seats in total.
• (c) The theatre’s total seating capacity is $\boxed{1725}$ seats.