Rd Chapter 9 Arithmetic Progressions Ex 9.3 Solutions
Question - 1 : - For the following arithmetic progressions write the first term a and the common difference d :
(i) -5, -1, 3, 7, …………
(ii) 1/5 , 3/5 , 5/5 , 7/5 , ……
(iii) 0.3, 0.55, 0.80, 1.05, …………
(iv) -1.1, -3.1, -5.1, -7.1, …………..
Answer - 1 : - (i) -5, -1, 3, 7, …………
Question - 2 : - Write the arithmetic progression when first term a and common difference d are as follows:
(i) a = 4, d = -3
(ii) a = -1, d= 1/2
(iii) a = -1.5, d = -0.5
Answer - 2 : -
(i) First term (a) = 4
and common difference (d) = -3
Second term = a + d = 4 – 3 = 1
Third term = a + 2d = 4 + 2 x (-3) = 4 – 6 = -2
Fourth term = a + 3d = 4 + 3 (-3) = 4 – 9 = -5
Fifth term = a + 4d = 4 + 4 (-3) = 4 – 12 = -8
AP will be 4, 1, -2, -5, -8, ……….
Question - 3 : - In which of the following situations, the sequence of numbers formed will form an A.P?
(i) The cost of digging a well for the first metre is ₹ 150 and rises by ₹ 20 for each succeeding metre.
(ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of the remaining in the cylinder.
(iii) Divya deposited ₹ 1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, …, and so on. [NCERT Exemplar]
Answer - 3 : -
(i) Cost of digging a well for the first metre = ₹ 150
Cost for the second metre = ₹ 150 + ₹ 20 = ₹ 170
Cost for the third metre = ₹ 170 + ₹ 20 = ₹ 190
Cost for the fourth metre = ₹ 190 + ₹ 20 = ₹ 210
The sequence will be (In rupees)
150, 170, 190, 210, ………..
Which is an A.P.
Whose = 150 and d = 20
(ii) Let air present in the cylinder = 1
(iii)Amount at the end of the 1st year = ₹ 1100
Amount at the end of the 2nd year = ₹ 1210
Amount at the end of 3rd year = ₹ 1331 and so on.
So, the amount (in ₹) at the end of 1st year, 2nd year, 3rd year, … are
1100, 1210, 1331, …….
Here, a2 – a1 = 110
a3 – a2 = 121
As, a2 – a1 ≠ a3 – a2,it does not form an AP
Find the common difference and write the next four terms of each of the following arithmetic progressions :
(i) 1, -2, -5, -8, ……..
(ii) 0, -3, -6, -9, ……
(iii) -1, 1/4 , 3/2 , ……..
(iv) -1, – 5/6 , – 2/3 , ………..
Answer - 4 : -
Question - 5 : - Prove that no matter what the real numbers a and b are, the sequence with nth term a + nb is always an A.P. What is the common difference ?
Answer - 5 : -
an = a + nb
Let n= 1, 2, 3, 4, 5, ……….
a1 = a + b
a2 = a + 2b
a3 = a + 3b
a4 = a + 4b
a5 = a + 5b
We see that it is an A.P. whose common difference is b and a for any real valueof a and b
as a2 – a1 = a + 2b – a – b = b
a3 – a2 = a + 3b – a – 2b = b
a4 – a3 = a + 4b – a – 3b = b
and a5 – a4 = a + 5b – a – 4b = b
Question - 6 : - Find out which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
Answer - 6 : -
Question - 7 : - Find the common difference of the A.P. and write the next two terms :
(i) 51, 59, 67, 75, …….
(ii) 75, 67, 59, 51, ………
(iii) 1.8, 2.0, 2.2, 2.4, …….
(iv) 0, 1/4 , 1/2 , 3/4 , ………..
(v) 119, 136, 153, 170, ………..
Answer - 7 : -