PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4

PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

Question 1.
Which term of the A.P. 121, 117, 113, …………. is its first negative term?
Solution:
Given A.P is 121, 117, 113, …
Here a = T₁ = 121 ;T₂ = 117; T₃ = 113
d = T₂ – T₁ = 117 – 121 = – 4
Using formula, T_n = a + (n – 1) d
T_n = 121 + (n – 1) (- 4)
= 121 – 4n + 4
= 125 – 4n.
According to question :—
T_n < 0
or 125 – 4n < 0
or 125 < 4n or 4n > 125.
or n > 125/4
or n > 31 1/4.
But n must be integer, for first negative term.
∴ n = 32.
Hence, 32nd term be the first negative term of given A.P.

Question 2.
The sum of the third and the seventh term of an A.P. is 6 and their product ¡s 8. Find the sum of first sixteer
terms of an A.P.
Solution:
Let ‘a’ and ‘d’ be the first term and common diftèrence of given A.P.
According to 1st condition
T₃ + T₇ = 6
[a + (3 – 1)d] + [a + (7 – 1) d] = 6
∵ [T_n = a + (n – 1) d]
or a + 2d + a + 6d = 6
or 2a + 8d = 6
or a + 4d = 3 …………….(1)
According to 2nd condition
T₃ (T₇) = 8
[a + (3 – 1) d] [a + (7 – 1)d] = 8
∵ [T_n = a + (n – 1) d]
or (a + 2d) (a + 6d) = 8
or [3 – 4d + 2d] [3 – 4d + 6d] = 8
[Using (1), a = 3 – 4d]
or (3 – 2d) (3 + 2d) = 8
or 9 – 4d² = 8
or 4d² = 98
or d² = 1/4
d = ± 1/2

Question 3.
A ladder has rungs 25 cm apart (see fig.) The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2 latex]\frac{1}{2}[/latex] m apart, what is the length of the wood required for the rungs?

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 – S₁]
Solution:
Let ‘x’ denotes the number of any house.
Here a = T₁ = 1 ;d = 1
According to question,

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