Question

The seating for an outdoor stage is arranged such that there are 11 seats in the first row. For each additional row after the first row, there are 3 more seats than there are in the previous row. If there are 30 rows altogether, how many seats are there in all? Select one: O

A. 1470
B. 1635
C. 2940
D. 3270

Answer 1

Answer:

B. 1635

Step-by-step explanation:

We have been given that the seating for an outdoor stage is arranged such that there are 11 seats in the first row. For each additional row after the first row, there are 3 more seats than there are in the previous row.

We can see that the seating order is in form of arithmetic sequence, whose first term is 11 and common difference is 3.

Since there are 30 rows altogether, so we need to find the sum of 30 first terms of the sequence using sum formula.

$S_n=\frac{n}{2}[2a+(n-1)d]$, where,

$S_n=$ Sum of n terms,

n = Number of terms,

a = First term,

d = Common difference.

Upon substituting our given value is above formula, we will get:

$S_n=\frac{30}{2}[2(11)+(30-1)3]$

$S_n=15[22+(29)3]$

$S_n=15[22+87]$

$S_n=15[109]$

$S_n=1635$

Therefore, there are 1635 seats in all and option B is the correct choice.

Answer 2

Answer: the total number of seats is 1635

Step-by-step explanation:

For each additional row after the first row, there are 3 more seats than there are in the previous row. This means that the seats in each row is increasing in arithmetic progression with a common difference of 3. The formula for determining the sum of n terms of an arithmetic sequence is expressed as

Sn = n/2[2a + (n - 1)d]

Where

n represents the number of terms in the sequence.

a represents the first term,

d represents the common difference.

From the information given,

a = 11

n = 30

d = 3

Therefore,

S30 = 30/2[2×11 + (30 - 1)3]

S30 = 15[22 + 29×3]

S30 = 15 × 109= 1635