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.

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.

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.

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