# Question 11 and 12, Exercise 4.2

Solutions of Question 11 and 12 of Exercise 4.2 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 11

If Rs. $1000$ is saved on August 1, Rs. $3000$ on August $2$, Rs. $5000$ on August $3$ and so on, How much is saved till August $20$ ?

** Solution. **

The sequence of the saved money is $$1000, 3000, 5000, \dots, \text{ upto 20 terms}.$$ This is in A.P with $a_1 = 1000$, $d=3000-1000=2000$, $S_20=?$ Since $$S_n =\frac{n}{2}[2a_1+(n-1)d],$$ implies \begin{align*} S_{20} &= \frac{20}{2}[2(1000)+(20-1)2000]\\ &= 10 [2000+(19)2000] \\ &= 10 [2000+38000] \\ &=10(40000)\\ &=400000. \end{align*} Hence, the total savings till August $20$ is Rs. $400,000.$

## Question 12

A gardener is making a triangular planting, with $35$ plants in the first row, $31$ in the second row, $27$ in the third row and so on. If the pattern is consistent, how many plants will there be in the eighth row?

** Solution. **

Given sequence of the plants in rows:
$$35, 31, 27, \ldots, a_8$$
This is an A.P with $a_1=35$, $d=31-35=-4$ and $a_8=?$.

We have
$$a_n=a_1+(n-1)d.$$
Thus, we have
\begin{align*}
a_8 &= 35 + (7)(-4) \\
&=35-28\\
&=7. \end{align*}
Hence $a_8=7$, that is, $7$ plants will be there in the 8th row.

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