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Chapter 9: Arithmetic Progressions Exercise 9.2 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, .......... Solution: We know that if a is the first term and d is the common difference, the arithmetic progression is a, a + d, a + 2d + a + 3d, ............ (i) – 5, –1, 3, 7, ............ Given arithmetic series is – 5, –1, 3, 7 ............. This is in the form of a, a + d, a + 2d + a + 3d, ............ by comparing these two a = – 5, a + d = 1, a + 2d = 3, a + 3d = 7, First term (a) = – 5 By subtracting second and first term, we get (a + d) - (a) = d -1 - (- 5) = d 4 = d Common difference (d) = 4. (ii) 1/5, 3/5, 5/5, 7/5, ............. Given arithmetic series is, 1/5, 3/5, 5/5, 7/5, ............... This is in the form of 1/5, 2/5, 5/5, 7/5, ........... a, a + d, a + 2d, a + 3d, By comparing this two, we get a = 1/5, a + d = 3/5, a + 2d = 5/5, a + 3d = 7/5 First term cos = 1/5 By subtracting first term from second term, we get d = (a + d)-(a) d = 3/5 - 1/5 d = 2/5 common difference (d) = 2/5 (iii) 0.3, 0.55, 0.80, 1.05, ............ Given arithmetic series, 0.3, 0.55, 0.80, 1.05, .......... General arithmetic series a, a + d, a + 2d, a + 3d, By comparing, a = 0.3, a + d = 0.55, a + 2d = 0.80, a + 3d = 1.05 First term (a) = 0.3. By subtracting first term from second term. We get d = (a + d) - (a) d = 0.55 - 0.3 d = 0.25 Common difference (d) = 0.25 (iv) –1.1, – 3.1, – 5.1, –7.1, ............. General series is a, a + d, a + 2d, a + 3d, ................ By comparing this two, we get a = –1.1, a + d = –3.1, a + 2d = –5.1, a + 3d = –71 First term (a) = –1.1 Common difference (d) = (a + d) - (a) = -3.1 – ( – 1.1) Common difference (d) = – 2 Question: 2 Write the arithmetic progressions write 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 Solution: We know that, if first term (a) = a and common difference = d, then the arithmetic series is, a, a + d, a + 2d, a + 3d, (i) a = 4, d = -3 Given first term (a) = 4 Common difference (d) = -3 Then arithmetic progression is, a, a + d, a + 2d, a + 3d, .............. ⟹ 4, 4 - 3, a + 2(-3), 4 + 3(-3), .............. ⟹ 4, 1, – 2, – 5, – 8 ................ (ii) a = -1, d = 1/2 Given, First term (a) = -1 Common difference (d) = 1/2 Then arithmetic progression is, ⟹ a, a + d, a + 2d, a + 3d, ⟹ -1, -1 + 1/2, -1, 2½, -1 + 3½, ........... ⟹ -1, -1/2, 0, 1/2 (iii) a = –1.5, d = – 0.5 Given First term (a) = –1.5 Common difference (d) = – 0 5 Then arithmetic progression is ⟹ a, a + d, a + 2d, a + 3d, ............... ⟹ -1.5, -1.5, -0.5, –1.5 + 2(– 0.5), –1.5 + 3(– 0.5) ⟹ – 1.5, – 2, – 2.5, – 3, ................ Then required progression is -1.5, – 2, – 2.5, – 3, ............... 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 Rs 150 and rises by Its 20 for each succeeding metre. (ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of their remaining in the cylinder. Solution: (i) Given, Cost of digging a well for the first meter (c1) = Rs.150. Cost rises by Rs.20 for each succeeding meter Then, Cost of digging for the second meter (c2) = Rs.150 + Rs 20 = Rs 170 Cost of digging for the third meter (c3) = Rs.170 + Rs 20 = Rs 210 Thus, costs of digging a well for different lengths are 150, 170, 190, 210, ........... Clearly, this series is in A∙P. With first term (a) = 150, common difference (d) = 20 (ii) Given Let the initial volume of air in a cylinder be V liters each time 3th/4 of air in a remaining 1.e 1 -1/4 First time, the air in cylinder is 3/4 V. Second time, the air in cylinder is 3/4 V. Third time, the air in cylinder is (3/4)2 V. Therefore, series is V, 3/4 V, (3/4)2 V,(3/4)3 V, ........ Question: 4 Show that the sequence defined by an = 5n - 7 is an A.P., find its common difference. Solution: Given sequence is an = 5n - 7 nth term of given sequence (an) = 5n -7 (n + 1)th term of given sequence (an + 1) - an = (5n - 2) - (5n - 7) = 5 ∴ d = 5 Question: 5 Show that the sequence defined by an = 3n2 - 5 is not an A.P. Solution: Given sequence is, an = 3n2 - 5. nth term of given sequence (an) = 3n2 - 5. (n + 1)th term of given sequence (an + 1) = 3(n + 1)2 - 5 = 3(n2 + 12 + 2n.1) - 5 = 3n2 + 6n -2 ∴ The common difference (d) = an + 1 - an d = (3n2 + 6n - 2) - (3n2 - 5) = 3a2 + 6n - 2 - 3n2 + 5 = 6n + 3 Common difference (d) depends on 'n' value ∴ given sequence is not in A.P.
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, ..........
We know that if a is the first term and d is the common difference, the arithmetic progression is a, a + d, a + 2d + a + 3d, ............
(i) – 5, –1, 3, 7, ............
Given arithmetic series is – 5, –1, 3, 7 .............
This is in the form of a, a + d, a + 2d + a + 3d, ............ by comparing these two a = – 5, a + d = 1, a + 2d = 3, a + 3d = 7,
First term (a) = – 5
By subtracting second and first term, we get
(a + d) - (a) = d
-1 - (- 5) = d
4 = d
Common difference (d) = 4.
(ii) 1/5, 3/5, 5/5, 7/5, .............
Given arithmetic series is,
1/5, 3/5, 5/5, 7/5, ...............
This is in the form of 1/5, 2/5, 5/5, 7/5, ........... a, a + d, a + 2d, a + 3d,
By comparing this two, we get
a = 1/5, a + d = 3/5, a + 2d = 5/5, a + 3d = 7/5
First term cos = 1/5
By subtracting first term from second term, we get
d = (a + d)-(a)
d = 3/5 - 1/5
d = 2/5
common difference (d) = 2/5
(iii) 0.3, 0.55, 0.80, 1.05, ............
Given arithmetic series,
0.3, 0.55, 0.80, 1.05, ..........
General arithmetic series
a, a + d, a + 2d, a + 3d,
By comparing,
a = 0.3, a + d = 0.55, a + 2d = 0.80, a + 3d = 1.05
First term (a) = 0.3.
By subtracting first term from second term. We get
d = (a + d) - (a)
d = 0.55 - 0.3
d = 0.25
Common difference (d) = 0.25
(iv) –1.1, – 3.1, – 5.1, –7.1, .............
General series is
a, a + d, a + 2d, a + 3d, ................
a = –1.1, a + d = –3.1, a + 2d = –5.1, a + 3d = –71
First term (a) = –1.1
Common difference (d) = (a + d) - (a)
= -3.1 – ( – 1.1)
Common difference (d) = – 2
Write the arithmetic progressions write 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
We know that, if first term (a) = a and common difference = d, then the arithmetic series is, a, a + d, a + 2d, a + 3d,
(i) a = 4, d = -3
Given first term (a) = 4
Common difference (d) = -3
Then arithmetic progression is, a, a + d, a + 2d, a + 3d, ..............
⟹ 4, 4 - 3, a + 2(-3), 4 + 3(-3), ..............
⟹ 4, 1, – 2, – 5, – 8 ................
(ii) a = -1, d = 1/2
Given, First term (a) = -1
Common difference (d) = 1/2
Then arithmetic progression is,
⟹ a, a + d, a + 2d, a + 3d,
⟹ -1, -1 + 1/2, -1, 2½, -1 + 3½, ...........
⟹ -1, -1/2, 0, 1/2
Given First term (a) = –1.5
Common difference (d) = – 0 5
Then arithmetic progression is
⟹ a, a + d, a + 2d, a + 3d, ...............
⟹ -1.5, -1.5, -0.5, –1.5 + 2(– 0.5), –1.5 + 3(– 0.5)
⟹ – 1.5, – 2, – 2.5, – 3, ................
Then required progression is
-1.5, – 2, – 2.5, – 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 Rs 150 and rises by Its 20 for each succeeding metre. (ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of their remaining in the cylinder.
(i) Given,
Cost of digging a well for the first meter (c1) = Rs.150.
Cost rises by Rs.20 for each succeeding meter
Then,
Cost of digging for the second meter (c2) = Rs.150 + Rs 20 = Rs 170
Cost of digging for the third meter (c3) = Rs.170 + Rs 20 = Rs 210
Thus, costs of digging a well for different lengths are 150, 170, 190, 210, ...........
Clearly, this series is in A∙P.
With first term (a) = 150, common difference (d) = 20
(ii) Given
Let the initial volume of air in a cylinder be V liters each time 3th/4 of air in a remaining 1.e
1 -1/4
First time, the air in cylinder is 3/4 V.
Second time, the air in cylinder is 3/4 V.
Third time, the air in cylinder is (3/4)2 V.
Therefore, series is V, 3/4 V, (3/4)2 V,(3/4)3 V, ........
Show that the sequence defined by an = 5n - 7 is an A.P., find its common difference.
Given sequence is
an = 5n - 7
nth term of given sequence (an) = 5n -7
(n + 1)th term of given sequence (an + 1) - an
= (5n - 2) - (5n - 7) = 5
∴ d = 5
Show that the sequence defined by an = 3n2 - 5 is not an A.P.
Given sequence is,
an = 3n2 - 5.
nth term of given sequence (an) = 3n2 - 5.
(n + 1)th term of given sequence (an + 1) = 3(n + 1)2 - 5
= 3(n2 + 12 + 2n.1) - 5
= 3n2 + 6n -2
∴ The common difference (d) = an + 1 - an
d = (3n2 + 6n - 2) - (3n2 - 5)
= 3a2 + 6n - 2 - 3n2 + 5
= 6n + 3
Common difference (d) depends on 'n' value
∴ given sequence is not in A.P.
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