Question 9 and 10, Exercise 4.2

Solutions of Question 9 and 10 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.

If $\dfrac{1}{a}, b, \dfrac{1}{c}$ are in A.P. Show that the common difference is $\dfrac{a-c}{2 a c}$.

Solution.

Since $\dfrac{1}{a}, b, \dfrac{1}{c}$ are in A.P.
$$\begin{align*} d&=b-\frac{1}{a}\cdots (i)\\ \end{align*}$$ Also $$\begin{align*} d&=\frac{1}{c}-b \cdots (ii) \end{align*}$$ Comparing (i) and (ii) we have
$$\begin{align*} b-\frac{1}{a}&=\frac{1}{c}-b\\ b+b&=\frac{1}{c}+\frac{1}{a}\\ 2b&= \frac{a+c}{ac}\\ b&= \frac{a+c}{2ac} \end{align*}$$ Putting the value of $b$ in (i), we have
$$\begin{align*} d&=\frac{a+c}{2ac}-\frac{1}{a}\\ &=\frac{(a+c)-2c}{2ac}\\ &=\frac{a+c-2c}{2ac}\\ &=\frac{a-c}{2ac} \end{align*}$$ As required.

During a free fall, a sky diver falls 16 feet in the first second, 48 feet in the 2nd second and 80 feet in the third second. If he continues to fall at this rate, how many feet will he fall during the 8 th second?

Solution.

By the given data, we have $$\begin{align*}16, 48, 80, \cdots \end{align*}$$ This is an A.P with $a_1=16$ and $d=48-16=32$. We have $$a_n = a_1 + (n - 1) d.$$ This gives $$\begin{align*} a_8 &= 16 + (7)(32)\\ &= 16 + (7)(32) = 240. \end{align*}$$ Hence the skydiver will fall $240$ feet during the 8th second.

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