# Question 23 and 24, Exercise 4.3

Solutions of Question 23 and 24 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

## Question 23

A formation of a marching band has 14 marches in the front row, 16 in the second row, 18 in the third row and so on, for 25 rows. How many marchers are in the last row? How many marchers are there altogether?

** Solution. **

From the statement, we have the following series:
$$ 14+16+18+...+a_{25}.$$
This is arithmetic series with $a_1=14$, $d=16-14=2$, $n=25$. We have to find $a_25$ and $S_25$.

As
\begin{align}
a_n&=a_1+(n-1)d\\
\implies a_{25}&= 14+(25-1)(2)\\
&=62.
\end{align}
Now
\begin{align}
S_n&=\frac{n}{2}[a_1+a_n]\\
\implies S_{25}& =\frac{25}{2}[14+62]\\
& =25 \times 38\\
& =950.
\end{align}
Hence $62$ marchers in last row and $950$ marchers are there altogether.

## Question 24

How many poles will be in a pile of telephone poles if there are 50 in the first layer, 49 in the second and so on, until there are 6 in the last layer?

** Solution. **

From the statement, we have the following series:
$$ 50+49+48+...+6.$$
This is arithmetic series with $a_1=50$, $d=49-50=-1$, $a_n=6$. We have to find $n$ and $S_n$.

As
\begin{align}
a_n&=a_1+(n-1)d\\
\implies 6&= 50+(n-1)(-1)\\
\implies 6&= 50-n+1\\
\implies n&= 51-6=45\\
\end{align}
Now
\begin{align}
S_n&=\frac{n}{2}[a_1+a_n]\\
\implies S_{45}& =\frac{45}{2}[50+6]\\
& =frac{45}{2} \times 56\\
& =1260.
\end{align}
Hence there are 1260 poles in the pile.

### Go to