ML Aggarwal Solution Class 10 Chapter 9 Arithmetic and Geometric Progressions Exercise 9.1
Exercise 9.1
Question 1
For the following A.P.s, write the first term a and the common difference d:
(i) 3, 1, – 1, – 3, …
(iii) – 3.2, – 3, – 2.8, – 2.6, …
Here first common term $(a)=\frac{1}{3}$
and common difference $(d)=$
$\frac{5}{3}-\frac{1}{3}=\frac{4}{3}, \frac{9}{3}-\frac{5}{3}=\frac{4}{3}, \ldots$
$=\frac{4}{3}$
(iii) -3.2,-3,-2.8,-2.6, ....
Here first term (a)=-3.2
and common difference (d)
=-3-(-3.2)=-3+3.2=0.2
=(d)=0.2
Question 2
Write first four terms of the A.P., when the first term a and the common difference d are given as follows :
(i) a = 10, d = 10
(ii) a = – 2, d = 0
(iii) a = 4, d = – 3
Sol :
(i) a = 10, d = 10
∴ A.P. = 10, 20, 30, 40, …
(ii) a = -2, d = 0
∴ A.P. = -2, -2, -2, -2, ….
(iv) $a=\frac{1}{2}, d=-\frac{1}{6}$
A.P. is $\frac{1}{2},\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{2}{6}$
$=\frac{2}{6}-\frac{1}{6}=\frac{1}{6}, \ldots$
A.P. $=\frac{1}{2}, \frac{2}{6}, \frac{1}{6}, 0, \ldots$
$=\frac{1}{2}, \frac{1}{3}, \frac{1}{6}, 0, \ldots$
Question 3
Which of the following lists of numbers form an A.P.? If they form an A.P., find the common difference d and write the next three terms :
(i) 4, 10, 16, 22,…
(ii) – 2, 2, – 2, 2,…..
(iii) 2, 4, 8, 16,….
(v) – 10, – 6, – 2, 2,….
(vi) 1², 3², 5², 7²,….
(vii) 1, 3, 9, 27,….
(viii) √2, √8, √18, √32,….
(ix) 3, 3 + √2, 3 + √2, 3 + 3√2,…..
(x) √3, √6, √9, √12,……
(xi) a, 2a, 3a, 4a,…….
(xii) a, 2a + 1, 3a + 2, 4a + 3,….
Sol :
(i) 4, 10, 16, 22,…
Here a = 4, d = 10 – 4 = 6, 16 – 10 = 6, 22 – 16 = 6
∵ common difference is same
∵ It is in A.P
and next three terms are 28, 34, 40
(iii) 2,4,8,16,...
Here, $a=2$
d=4-2=2,8-4=4,16-8=8
$\therefore$ Common difference is not same.
$\therefore$ It is not an A.P.
(iv) $2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$
Here a=2
$d=\frac{5}{2}-2=\frac{1}{2}$
$3-\frac{5}{2}=\frac{1}{2}$
$\frac{7}{2}-3=\frac{1}{2}$
$\because$ Common difference is same.
$\therefore$ It is an A.P.
and next three terms are $4, \frac{9}{2}, 5$
(v) $-10,-6,-2,2, \ldots$
Here, first term $(a)=-10$
$d=-6-(-10)=-6+10=4$
$-2-(-6)=-2+6=4$
$2-(-2)=2+2=4$
∴Common difference is same.
∴It is an A.P.
and next three terms are 6,10,14,20
(vi)
$1^{2}, 3^{2}, 5^{2}, 7^{2}, \ldots$
=1,9,25,49, ...
Here, first term $(a)=1^{2}=1$
d=9-1=8
25-9=16
49-25=24
∵Common difference is not same.
∴It is not an A.P.
(vii) 1,3,9,27,...
Here , first term (a)=1
d=3-1=2
9-3=6
27-9=18
∵Common difference is not same.
∴It is not an A.P.
(viii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$
⇒$\sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, 4 \sqrt{2}, \ldots$
Here, first term $(a)=\sqrt{2}$
and common difference $(d)$
$=2 \sqrt{2}-\sqrt{2}=\sqrt{2}$
$=3 \sqrt{2}-2 \sqrt{2}=\sqrt{2}$
$=4 \sqrt{2}-3 \sqrt{2}=\sqrt{2}$
$\because$ The common difference is same
$\therefore$ It is an A.P.
$\sqrt{50}, \sqrt{72}, \sqrt{98}, \ldots$
(ix) $3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots$
Here, first term $(a)=3$
and $d=3+\sqrt{2}-2=\sqrt{2}$
$3+2 \sqrt{2}-3-\sqrt{2}=\sqrt{2}$
$3+3 \sqrt{2}-3+2 \sqrt{2}=\sqrt{2}$
∵Common difference is same.
∴It is an A.P.
and next three terms are
$3+4 \sqrt{2}, 3+5 \sqrt{2}, 3+6 \sqrt{2}, \ldots$
(x) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$
Here, $a=\sqrt{3}$
$d=\sqrt{6}-\sqrt{3}=\sqrt{3} \times \sqrt{2}-\sqrt{3}$
$=\sqrt{3}(\sqrt{2}-1)$
$=\sqrt{9}-\sqrt{6}=3-\sqrt{2} \sqrt{3}=\sqrt{3}(\sqrt{3}-\sqrt{2})$
∴Common difference is not same.
∴It is not an A.P.
(xi) a, 2 a, 3 a, 4 a, ...
Here first term (a)=a
Common difference (d)=2 a-a=a
3a-2a=a
4a-3 a=a
∵The common difference is same.
∴It is an A.P.
and next three terms are
5a, 6a, 7a
(xii) a,2a+1, 3a+2, 4a+3...
Here, first term (a)=a
and common difference (d)
=2a+1-a=a+1
3a+2-2a-1=a+1
4a+3-3a-2=a+1
∵The common difference is same.
∴It is an A.P.
and next three terms are
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