Question 9 and 10, Exercise 4.3

Solutions of Question 9 and 10 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.

Find the sum of the odd numbers from $1$ to $99$.

Solution.

Solution. Sum of the odd numbers from $1$ to $99$ is $$1+3+5+...+99 (50 \text{ terms}).$$ This is arithmetic series with: $a_{1}=1$, $n=50$, $d=3-1=2$.
Let $S_n$ represents sum of the arithmetic series. Then $$\begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \implies S_{50}&=\frac{50}{2}[2(1)+(50-1)(2)]\\ &=25\times [2+98]\\ &=2500. \end{align}$$ Hence the sum of the odd numbers from $1$ to $99$ is $2500$.

Find the sum of all multiples of 4 that are between $14$ and $523$.

Solution.

Sum of all multiples of 4 that are between $14$ and $523$. $$16+20+24+...+520.$$ This is arithmetic series with: $a_{1}=16$, $d=20-16=4$, $a_n=520$, $n=?$.
We have $$\begin{align} & a_n=a_1+(n-1)d \\ \implies & 520=16+(n-1)(4) \\ \implies & 520=16+4n-4 \\ \implies & 4n=520-16+4 \\ \implies & 4n = 508\\ \implies & n = 127. \end{align}$$ Let $S_n$ represents sum of the arithmetic series. Then $$\begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{127}&=\frac{127}{2}[16+520]\\ &=\frac{127}{2}\times 536\\ &=34036. \end{align}$$ Hence the required sum is $34036$.

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