TS 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise

Question 1.
Which term of the AP:
121, 117, 113,…, is the first negative term?
[Hint: Find n for a_n < 0]
Solution:
Given A.P. : 121, 117, 113,……….
a = 121 and d = a₂ – a₁
= 117 – 121 = -4
Let the n^th term be the first negative term of the given G.P
Then, a_n = 0
⇒ a_n + (n – 1)d < 0
⇒ 121 + (n – 1)(-4) < 0
⇒ 121 – 4n + 4 < 0
⇒ 121 – 4n < 0
⇒ -4n < -125 ⇒ 4n > 125
⇒ n > \(\frac{125}{4}\)
n > 31.25
∴ When n = 32, the term becomes negative, (or) 32^nd term is the first negative term of the given A.P.

Question 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 theAP. ?
Solution:
Given in an A.P : t₃ + t₇ = 6
t₃ + t₇ = 8
a_n = t_n = a + (n – 1) d
Let the terms of the A.R be
(a – 4d), (a – 3d), (a – 2d), (a – d), a(a + d), (a + 2d), (a + 3d),….
Then a₃ = a – 2d; a₇ = a + 2d
So a₃ + a₇ = a – 2d + a + 2d = 6
⇒ 2a = 6 (or) a = 3
Also, a₃ . a₇ = (a – 2d) (a + 2d)
⇒ a² – 4d² = 8
⇒ 3² – 4d² = 8
⇒ 9 – 4d² = 8
⇒ 4d² = 9 – 8
⇒ d² = \(\frac{1}{4}\)
∴ d = ± \(\frac{1}{2}\)
Taking d = \(\frac{1}{2}\)

Question 3.
A ladder has rungs 25 cm apart. 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\(\frac{1}{2}\) m apart, what is the length of the wood required for the rungs ?
[Hint : Number of rungs = \(\frac{250}{25}\) + 1]

Solution:
Given : A ladder with rungs separated by a distance between the rungs = 2\(\frac{1}{2}\)m = 250 cm
∴ Number of rungs = \(\frac{250}{25}\) + 1
= 10 + 1 = 11
Length of the bottom and top rungs = 45 cm and 25 cm
where a = 45; last term l = a₁₁ = 25 and d = 2 cm
∴ S_n = (a + l) = \(\frac{11}{2}\)(45 + 25)
= \(\frac{11}{2}\) × 70
= 35 × 11 = 385 cm
∴ Length of wood required for total rungs = 385 cm.

Question 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. And find this value of x.
[Hint : S_x-1 = S₄₉ – S_x]
Solution:
Given: Houses with numbers from 1 to 49.
x is a number x such that,
S_x-1 = S₄₉ – S_x
We know that, S_x = \(\frac{x}{2}\)[2a + (x – 1)d]
Where a = 1; d = a₂ – a₁ = 2 – 1 = 1

⇒ x(x – 1) + x(x + 1) = 2 × 1225
⇒ x + x² + x = 2450
⇒ 2x² = 2450
x² = \(\frac{2450}{2}\) = 1225
⇒ x = \(\sqrt{1225}\) = ±35
⇒ x = 35
(∵ x is a counting number)

Question 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 \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\) m. (See Fig.). Calculate the total volume of concrete required to build the terrace.
[Hint : Volume of concrete required to build the first step = \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m³]

Solution:
Given No. of steps = 15
i.e., n = 15
Volume of concrete required to build the second step = (\(\frac{1}{4}\) + \(\frac{1}{4}\)) × \(\frac{1}{2}\) × 50
= \(\frac{1}{2}\) × \(\frac{1}{2}\) × 50
= \(\frac{25}{2}\) m³

∴ Total volume of the concrete required
= 750 m³

Question 6.
150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped from the work in the second day. Four workers dropped in third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
[Let the no.of days to finish the work is ‘x’ then
150x = \(\frac{x+8}{2}\)[2 × 150 + (x + 8 – 1)(-4)]
Solution:
Let the number of days to finish the work is x.
No. of workers engaged = 150
Given four workers dropped from the work in the second day.
Four workers dropped in the third day and so on.
∴ Common difference d = -4
8 more days to finish the work
n = x + 8
a = 150
S_n = 150x
\(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d] = 150x
\(\left(\frac{x+8}{2}\right)\)[2 × 150 + (x + 8 – 1)(-4)] = 150x
(x + 8)[150 + (x + 7)(-2)] = 150x
(x + 8) [150 – 2x – 14] = 150x
(x + 8)(136 – 2x) = 150x
136x – 2x² + 1088 – 16x = 150x
-2x² + 120x + 1088 – 150x = 0
-2x² – 30x + 1088 = 0
2x² + 30x – 1088 = 0
x² + 15x – 544 = 0
x² – 17x + 32x – 544 = 0
x(x – 17) + 32(x – 17) = 0
(x – 17) (x + 32) = 0
Either x – 17 = 0 or x + 32 = 0
x = 17 (or) x = -32
Number of days can’t be negative.
∴ x = 17
∴ Number of days to complete the work is
17 + 8 = 25

Question 7.
A machine costs ₹ 5,00,000. If the value depreciates 15% in the first year, 13\(\frac{1}{2}\) % in the second year, 12% in the third year and so on. What will be its value at the end of 10 years, when all the percent ages will be applied to the original cost?
Solution:
Given : Cost price of a machine = ₹ 5,00,000
Depreciation during the years

Sum of the depreciations =
15 + 13\(\frac{1}{2}\) +12+……..+10 terms
a = 15; d = a₂ – a₁
= 13\(\frac{1}{2}\) – 15 = -1\(\frac{1}{2}\) = –\(\frac{3}{2}\)

∴ Cost after 10 years = (100 – 82.5)% of 5,00,000
= 17.5% of 5,00,000
= \(\frac{17.5 \times 500000}{100}\)
= Rs. 87,500
∴ The value at the end of 10 years will be Rs. 87,500.