# Question 13 & 14 Exercise 4.3

Solutions of Question 13 & 14 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 13

A theater has 40 rows with 20 seats in the first row, 23 in the second row, 26 in the third row and so forth. How many seats are there in the theater?

### Solution

\begin{align}\text{Total number of rows}& n=40,\\
\text{Seats in a first row} a_1&=20\\
\text{Seat in a second row} a_2&=23\\
\text{Seats in third row} a_3&=26\end{align}

and so on upto 40 rows. Thus the sequence $20,23,26, \ldots$ is an arithmetic equence.

We have to find the total number of seats that are $S_{40}$.

We know by sum formula

$$S_n=\dfrac{n}{2} [{2} a_1+(n-1) d] \text {.}$$
That becomes

\begin{align}S_{40}&=\dfrac{40}{2}[2 \cdot 20+39(3)] \\
\Rightarrow S_{40}&=20[40+117] \\
\Rightarrow S_{40}&=20(157)=3140\end{align}
Thus the total seats in theater are 3140.

## Question 14

Insert enough arithmetic means between $1$ and $50$ so that the sum of the resulting series will be $459$ .

### Solution

We know that:

\begin{align}S_n&=\dfrac{n}{2}[a_1+a_n],\\
\text{ putting the given}\\
459&=\dfrac{n}{2}[1+50] \\
\Rightarrow n(51)&=918 \\
\Rightarrow n&=\dfrac{918}{51}=18.\end{align}
However this is including $1$ and $50$ as terms,

so therefore there would need to be $16$ terms between $1$ and $50$.

### Go To