Question

In a small church in rural pennsylvania, the first pew can fit only 6 people. the second pew can fit 8, and the third can fit 10 people. if this pattern continues, how many people can fit in the church if there are 15 pews, and how many people can fit into the last pew

Answer 1

An=a1+(n-1)d
A15=6+(15-1)2
A15=6+(14)2
A15=6+28
A15=34
Answer=34

Answer 2

Given, In a small church, the first pew can fit 6 people, second pew can fit 8 people, third pew can fit 10 people and so on.

We have to find how many people can fit in the church if there are 15 pews.

Here 6, 8, 10, ..... is in arithmetic series, where the common difference is (8-6) = (10-8) = 2.

To find total number of people, we have to find the sum of series for total numner of terms 15.

The formula to find the sum is

$S = \frac{n}{2}(2a+(n-1)d)$

Where, n = total number of terms, a = first term, d = common difference.

Here, n = 15, a = 6, d = 2

By substituting the values in the formula we will get,

$S =\frac{15}{2}(2(6)+(15-1)(2))$

$S= \frac{15}{2}(12+(15-1)(2))$

$S= \frac{15}{2}(12+(14)(2))$

$S = \frac{15}{2} (12+28)$

$S = \frac{15}{2}(40)$

$S = \frac{(15)(40)}{2}$

$S = \frac{600}{2}$

$S = 300$

So the total number of people that can fit in the church is 300.

Now we have to find the number of people that can fit into the last pew that is in the 15th pew.

We have to use the formula to find nth term.

The formula is $a_{n} = a+(n-1)d$

Here, n = 15, a = 6 and d =2.

By substituting the values we will get,

$a_{15} = 6+(15-1)(2)$

$a_{15}= 6+(14)(2)$

$a_{15} = 6+28$

$a_{15} = 34$

So we have got, 34 people can fit in the last pew that is in the 15th pew.

We have got the required answers here.