# Question 5 and 6, Exercise 4.3

Solutions of Question 5 and 6 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 5

Find the sum of series. $a_{1}=50$, $n=20$, $d=-4$.

Statement is logically incorrect.

Find the sum of arithmetic series with: $a_{1}=50$, $n=20$, $d=-4$.

** Solution. **
Given: $a_{1}=50$, $n=20$, $d=-4$.

Let $S_n$ represents sum of arithmetic series. Then
\begin{align}
S_n&=\frac{n}{2}[2a_1+(n-1)d] \\
\implies S_{20}&=\frac{20}{2}[2(50)+(20-1)(-4)]\\
&=10\times [100-76]\\
&=240.
\end{align}
Hence $S_{20}=240$.

## Question 6

Find the sum of series. $-3+(-7)+(-11)+\cdots +a_{10}$

** Solution. **
Given series is arithmetic series with $a_1=-3$, $d=-7-(-3)=-4$, $n=10$.

Let $S_n$ represents sum of arithmetic series. Then
\begin{align}
S_n&=\frac{n}{2}[2a_1+(n-1)d] \\
\implies S_{10}&=\frac{10}{2}[2(-3)+(10-1)(-4)]\\
&=5\times [-6-36]\\
&=-210.
\end{align}
Hence $S_{10}=-210$.

### Go to