# Arithmetic Progressions : Exercise 5.4 (Optional) (Mathematics NCERT Class 10th)

Q.1     Which term of the AP : 121, 117, 113....., is its first negative terms?
Sol.       121, 117, 113, ....
a = 121, d = 117 – 121 = – 4
${a_n} = a + \left( {n - 1} \right)d$
= 121 + (n – 1) × – 4
= 121 – 4n + 4 = 125 – 4n
For the first negative term
${a_n} < \,0\,\,\,\, \Rightarrow \,\,\,\,125\,\, - 4n < 0$
$\Rightarrow$ 125 < 4n
$\Rightarrow$ ${{125} \over 4} < n$
$\Rightarrow$ $31{1 \over 4} < n$
n is an integer and $n > 31{1 \over 4}$
$\Rightarrow$ The first negative term is 32nd term.

Q.2     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.
Sol.         Let the AP be a – 4 d, a – 3 d, a – 2 d, a – d, a, a + d, a + 2 d, a + 3 d, ....
Then,  ${a_3} = a - 2d,\,{a_7} = a - 2d$
$\Rightarrow$ ${a_3} + \,{a_7} = a - 2d + a - 2d = 6$
$\Rightarrow$ 2a = 6
$\Rightarrow$ a = 3 ... (1)

Also (a – 2d) (a + 2d) = 8
$\Rightarrow$ ${a^2} - 4{d^2} = 8$
$\Rightarrow$ $4{d^2} = {a^2} - 8$
$\Rightarrow$ $4{d^2} = {\left( 3 \right)^2} - 8 = 9 - 8 = 1$
$\Rightarrow$ ${d^2} = {1 \over 4}$
$\Rightarrow$ ${d^2} = \pm {1 \over 2}$
Taking $d = {1 \over 2}$
${S_{16}} = {{16} \over 2}\left[ {2 \times \left( {a - 4d} \right) + \left( {16 - 1} \right) \times d} \right]$
$= 8\left[ {2 \times \left( {3 - 4 \times {1 \over 2}} \right) + 15 \times {1 \over 2}} \right]$
$= 8\left[ {2 + {{15} \over 2}} \right] = 8 \times {{19} \over 2} = 76$
Taking $d = - {1 \over 2}$
${S_{16}} = {{16} \over 2}\left[ {2 \times \left( {a - 4d} \right) + \left( {16 - 1} \right) \times d} \right]$
$= 8\left[ {2 \times \left( {3 - 4 \times {1 \over 2}} \right) + 15 \times - {1 \over 2}} \right]$
$= 8\left[ {2 \times 5 - {{15} \over 2}} \right]$
$= 8\left[ {{{20 - 15} \over 2}} \right]$
$= 8 \times {5 \over 2} = 20$
Therefore, ${S_{16}} = 20,76$

Q.3      A ladder has rungs 25 cm apart (see figure). 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 $2{1 \over 2}$ m apart, what is the length of the wood required for the rungs? Sol.       Number of rungs, $n = {{2{1 \over 2}m} \over {25\,cm}}$
$= {{250\,cm} \over {25\,cm}} = 10$
So, there are 10 rungs
The length of the wood required for rungs = Sum of 10 rungs
$= {{10} \over 2}\left[ {25 + 45} \right]$
$= 5 \times 70 = 350\,cm$

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.
Sol.        Here a = 1, and d = 1
Therefore, ${S_{n - 1}} = {{x - 1} \over 2}\left[ {2 \times 1 + \left( {x - 1 - 1} \right) \times 1} \right]$e
$= {{x - 1} \over 2}\left( {2 + x - 2} \right) = {{\left( {x - 1} \right)\left( x \right)} \over 2}$
$= {{{x^2} - x} \over 2}$
${S_n} = {x \over 2}\left[ {2 \times 1 + \left( {x - 1} \right) \times 1} \right] = {x \over 2}\left( {x + 1} \right)$
${{{x^2} + x} \over 2}$
and, ${S_{49}} = {{49} \over 2}\left[ {2 \times 1 + \left( {49 - 1} \right) \times 1} \right]$
$= {{49} \over 2}\left[ {2 + 48} \right] = {{49} \over 2} \times 50$
$= 49 \times 25$
According to the question,
${S_{n - 1}} = {S_{49}} - {S_x}$
i.e., ${{{x^2} - x} \over 2} = 49 \times 25 - {{{x^2} + x} \over 2}$
$\Rightarrow$ ${{{x^2} - x} \over 2} + {{{x^2} + x} \over 2} = 49 \times 25$
$\Rightarrow$ ${{{x^2} - x + {x^2} + x} \over 2} = 49 \times 25$
$\Rightarrow$ ${x^2} = 49 \times 25$
$\Rightarrow$ $x = \pm \,\,7 \times 5$
Since x is a counting number , so taking positive square root, x = 7 × 5 = 35.

Q.5     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 ${1 \over 4}m$ and a tread of ${1 \over 2}m$ (see figure). Calculate the total volume of concrete required to build the terrace.

Sol.        Volume of concrete required to build the first step, second step, third step.... $\left( {in\,{m^3}} \right)$ are
${1 \over 4} \times {1 \over 2} \times 50,\left( {2 \times {1 \over 4}} \right) \times {1 \over 2} \times 50,\left( {3 \times {1 \over 4}} \right) \times {1 \over 2} \times 50,....$
i.e. ${{50} \over 8},2 \times {{50} \over 8},3 \times {{50} \over 8},....$
Therefore Total volume of concrete required.
$= {{50} \over 8} + 2 \times {{50} \over 8} + 3 \times {{50} \over 3} + ...$
$= {{50} \over 8}\left[ {1 + 2 + 3 + ...} \right]$
$= {{50} \over 8} \times {{15} \over 2}\left[ {2 \times 1 + \left[ {15 - 1)} \right] \times 1} \right]$ [Since, n = 15]
$= {{50} \over 8} \times {{15} \over 2} \times 16 = 750\,{m^3}$

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