====== Question 25 and 26, Exercise 4.3 ======
Solutions of Question 25 and 26 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 25=====
A family saves money in an arithmetic sequence: Rs. 6,000 in the first year, Rs. 70,000 in second year and so on, for 20 years. How much do they save in all?

** Solution. **

From the statement, we have the following series:
$$ 6000+70,000+...+a_{20}.$$
This is arithmetic series with $a_1=6,000$, $d=70,000-6,000=64,000$, $n=20$. We have to find $S_n$.\\
As
\begin{align}
S_n&=\frac{n}{2}[2a_1+(n-1)d]\\
\implies S_{20}& =\frac{20}{2}[2(6,000)+(20-1)(64,000)]\\
& =10 \times [12,000+1,216,000]\\
& =12,280,000.
\end{align}
Hence the family will save Rs. 12,280,000. GOOD 




=====Question 26=====
Mr. Saleem saves Rs. 500 on October 1, Rs. 550 on October 2, and Rs. 600 on October 3 and so on. How much is saved during October? (October has 31 days)

** Solution. **

From the statement, we have the following series:
$$ 500+550+600+...+a_{31}.$$
This is arithmetic series with $a_1=500$, $d=550-500=50$, $n=31$. We have to find $S_n$.\\
As
\begin{align}
S_n&=\frac{n}{2}[2a_1+(n-1)d]\\
\implies S_{31}& =\frac{31}{2}[2(500)+(31-1)(50)]\\
& =frac{31}{2} \times [1000+1500]\\
& =31 \times 1250\\
& =38750.
\end{align}
Hence Mr. Saleem will save Rs. 38,750. GOOD 

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