ARITHMETIC PROGRESSIONS-Exercise 5.4
Maths - Exercise
Q 1. Which term of the AP : 121, 117, 113, . . ., is its first negative term?
Solution:
AP: 121, 117, 113.... $$\begin{aligned}a=121\\ d=117-121\\ =-4\end{aligned}$$ Let a_n be the first negative number $$\scriptsize\begin{aligned}a_{n}=a+\left( n-1\right) d\\ a_{n} \lt 0\\ \\ \Rightarrow a+\left( n-1\right) d \lt &0\\ \Rightarrow 121+\left( n-1\right) \left( -4\right) \lt &0\\ \Rightarrow 121-\left( n-1\right) 4 \lt &0\\ \Rightarrow 121-4n+4\lt&0\\ \Rightarrow 125-4n \lt &0\\ \Rightarrow 4n \gt &125\\ n\gt&\dfrac{125}{4}\\ n\gt&31\dfrac{1}{4}\end{aligned}$$ therefore 32nd term will be first negative numberQ2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Solution:
Sum of third and seventh term of AP = 6 $$\scriptsize\begin{align}a_{3}+a_{7}&=6\\ \Rightarrow a_{1}+2d+a_{1}+6d&=6\\ \Rightarrow 2a_{1}+8d&=6\\ \Rightarrow a_{1}+4d&=3\\ \Rightarrow a_{1}&=3-4d\tag{1}\end{align}$$Product of third and seventh term of the AP
$$\scriptsize\begin{align}a_{3}\cdot a_{7}&=8\\ \Rightarrow \left( a_{1}+2d\right) \left( a_{1}+6d\right) &=8\\ \Rightarrow a_{1}^{2}+6a_{1}d+2a_{1}d+12d^{2}&=8\\ \Rightarrow a_{1}^{2}+8a_{1}d+12d^{2}&=8\tag{2}\end{align}$$Substituting value of \(a\), from equation-(1) to equation-(2)
$$\scriptsize\begin{aligned}\left( 3-4d\right) ^{2}+8\left( 3-4d\right) d+12d^{2}&=8\\ \left( 3-4d\right) ^{2}+8d\left( 3-4d\right) +12d^{2}&=8\\ 9+16d^{2}-24d+24d-32d^{2}+12d^{2}-8&=0\\ 9-8+16d^{2}-32a^{2}+12d^{2}&=0\\ 1+28d^{2}-32d^{2}&=0\\ -4d^{2}+1&=0\\ 4d^{2}&=1\\ d^{2}&=\dfrac{1}{4}\\ d&=\sqrt{\dfrac{1}{4}}\\ &=\dfrac{1}{2}\end{aligned}$$Substituting value of \(d\) in equation-(1)
$$\begin{aligned}a_{1}&=3-4d\\ &=3-4\times \dfrac{1}{2}\\ &=3-2\\ &=1\end{aligned}$$ $$a=1,\; d=\dfrac{1}{2}$$therefore sum of first 16 terms
$$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ S_{16}&=\dfrac{16}{2}\left[ 2\times 1+\left( 16-1\right) \times \dfrac{1}{2}\right] \\ &=8\left[ 2+15\times \dfrac{1}{2}\right] \\ &=8\left( 2+\dfrac{15}{2}\right) \\ &=8\times \dfrac{19}{2}\\ &=4\times 19\\ =76\end{aligned}$$Sum of first 16 terms \[\boxed{S_{16}=76}\]
Q A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \(2\frac{1}{2}\) apart, what the length of the wood required for the rungs?
Solution:
Distance of rungs = 25 cm
Length of of the ring at bottom = 45cm
Length of the rung on the top = 25cm
Distance between top and bottom rungs = 2.5 m = 250 cm
Therefore numbers of rungs
Length of the rungs decreases uniformly as' it goes up AP Formed by the rings
$$45,\ldots ,25$$ $$\scriptsize\begin{aligned}a_{1}&=45\\ a_{11}&=25\\ d&=25cm\\\\ S_{n}&=\dfrac{n}{2}\left[ a+l\right] \\ &=\dfrac{11}{2}\left[ 45+25\right] \\ &=\dfrac{11}{2}\times 70\\ &=11\times 35\\ &=385cm\end{aligned}$$Q 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of \(x\) such that the sum of the numbers of the houses preceding the house numbered \(x\) is equal to the sum of the numbers of the houses following it. Find this value of \(x\).
Solution:
Houses of Rows are numbered consecutively from 1- 49
Let there is a house numbered \(x\)
To show that sum of house number before \(x\) and after \(x\) are same
Sum of house number before \(x\) = \(S_{x-1}\)
Common difference \(d=1\)
First term \(a=1\)
And Sum of house after \(x\) =\(S_{49}-S_{x}\)
Q5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\) m and a tread of \(\frac{1}{2}\) m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace.
Solution:
Length of each terrace = 50 m tread = 1/2 m
Height = 1/4 m
volume of first terrace
Volume of 2nd terrace Height of 2nd terrace will become twice of first
$$\begin{aligned}H&=\dfrac{1}{4}\times 2=\dfrac{1}{2}\\\\ V_{2}&=50\times \dfrac{1}{4}\times 2\times \dfrac{1}{2}\\ &=\dfrac{50}{4}m^{3}\end{aligned}$$Similarly
$$\begin{aligned}V_{3}&=50\times \dfrac{1}{4}\times 3\times \dfrac{1}{2}\\ &=\dfrac{75}{4}m^{3}\end{aligned}$$AP formed by Volumes of steps of terrace
$$V_{1},V_{2},V_{3}....V_{15}$$ $$\begin{aligned}\dfrac{25}{7},\dfrac{50}{4},\dfrac{75}{4}\ldots\end{aligned}$$ $$\begin{aligned}a&=\dfrac{25}{4}\\ d&=\dfrac{50}{4}-\dfrac{25}{4}=\dfrac{25}{4}\\ n&=15\end{aligned}$$Total volume
$$\scriptsize\begin{aligned}S_{15}&=\dfrac{15}{2}\left[ 2\times \dfrac{25}{4}+\left( 15-1\right) \dfrac{25}{4}\right] \\ &=\dfrac{15}{2}\left[ \dfrac{25}{2}+14\times \dfrac{25}{4}\right] \\ &=\dfrac{15}{2}\left[ \dfrac{25}{2}+\dfrac{7\times 25}{2}\right] \\ &=\dfrac{15}{2}\left[ \dfrac{25+175}{2}\right] \\ &=\dfrac{15\times 200}{4}\\ &=15\times 50\\ &=750\ m^{2}\end{aligned}$$
