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Which of the following are APs? If they form an AP, find the common difference $d$ and write three more terms.
(i) \( 2,4,8,16, \ldots \)
(ii) \( 2, \frac{5}{2}, 3, \frac{7}{2}, \ldots \)
(iii) \( -1.2,-3.2,-5.2,-7.2, \ldots \)
(iv) \( -10,-6,-2,2, \ldots \)
(v) \( 3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots \)
(vi) \( 0.2,0.22,0.222,0.2222, \ldots \)
(vii) \( 0,-4,-8,-12, \ldots \)
(viii) \( -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots \)
(ix) \( 1,3,9,27, \ldots \)
(x) \( a, 2 a, 3 a, 4 a, \ldots \)
(xi) \( a, a^{2}, a^{3}, a^{4}, \ldots \)
(xii) \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots \)
(xiii) \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots \)
(xiv) \( 1^{2}, 3^{2}, 5^{2}, 7^{2}, \ldots \)
(xv) \( 1^{2}, 5^{2}, 7^{2}, 73, \ldots \)
To do:
We have to check whether the given sequences are in AP. If they are in AP, we have to find the common difference $d$ and write three more terms.
Solution:
(i) Given sequence is $2, 4, 8, 16, …….$
In the given sequence,
$a_1=2, a_2=4, a_3=8$
$a_2-a_1=4-2=2$
$a_3-a_2=8-4=4$
$a_2 - a_1 ≠ a_3 - a_2$
Therefore, the given sequence is not an AP.
(ii) Given sequence is $2, \frac{5}{2}, 3, \frac{7}{2}, …….$
In the given sequence,
$a_1=2, a_2=\frac{5}{2}, a_3=3$
$a_2-a_1=\frac{5}{2}-2=\frac{5-4}{2}=\frac{1}{2}$
$a_3-a_2=3-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}$
$a_2 - a_1 = a_3 - a_2$
$d=a_2 - a_1=\frac{1}{2}$
$a_5=a_4+d=\frac{7}{2}+\frac{1}{2}=\frac{7+1}{2}=\frac{8}{2}=4$
$a_6=a_5+d=4+\frac{1}{2}=\frac{8+1}{2}=\frac{9}{2}$
$a_7=a_6+d=\frac{9}{2}+\frac{1}{2}=\frac{9+1}{2}=\frac{10}{2}=5$
(iii) Given sequence is $-1.2, -3.2, -5.2, -7.2, ……$
In the given sequence,
$a_1=-1.2, a_2=-3.2, a_3=-5.2$
$a_2-a_1=-3.2-(-1.2)=-3.2+1.2=-2$
$a_3-a_2=-5.2-(-3.2)=-5.2+3.2=-2$
$a_2 - a_1 = a_3 - a_2$
$d=a_2 - a_1=-2$
$a_5=a_4+d=-7.2+(-2)=-7.2-2=-9.2$
$a_6=a_5+d=-9.2+(-2)=-9.2-2=-11.2$
$a_7=a_6+d=-11.2+(-2)=-11.2-2=-13.2$
(iv) Given sequence is $-10, -6, -2, 2, …..$
In the given sequence,
$a_1=-10.2, a_2=-6, a_3=-2$
$a_2-a_1=-6-(-10)=-6+10=4$
$a_3-a_2=-2-(-6)=-2+6=4$
$a_2 - a_1 = a_3 - a_2$
$d=a_2 - a_1=4$
$a_5=a_4+d=2+4=6$
$a_6=a_5+d=6+4=10$
$a_7=a_6+d=10+4=14$
(v) Given sequence is $3, 3 + \sqrt2, 3 + 2\sqrt2, 3 + 3\sqrt2, …..$
In the given sequence,
$a_1=3, a_2=3 + \sqrt2, a_3=3 + 2\sqrt2$
$a_2-a_1=3 + \sqrt2-3=\sqrt2$
$a_3-a_2=3 + 2\sqrt2-(3 + \sqrt2)=\sqrt2$
$a_2 - a_1 = a_3 - a_2$
$d=a_2 - a_1=\sqrt2$
$a_5=a_4+d=3 + 3\sqrt2+ \sqrt2=3 + 4\sqrt2$
$a_6=a_5+d=3 + 4\sqrt2 + \sqrt2=3 + 5\sqrt2$
$a_7=a_6+d=3 + 5\sqrt2+ \sqrt2=3 + 6\sqrt2$
(vi) Given sequence is $0.2, 0.22, 0.222, 0.2222, ……$
In the given sequence,
$a_1=0.2, a_2=0.22, a_3=0.222$
$a_2-a_1=0.22-0.2=0.02$
$a_3-a_2=0.222-0.22=0.002$
$a_2 - a_1 ≠ a_3 - a_2$
Therefore, the given sequence is not an AP.
(vii) Given sequence is $0, -4, -8, -12, …..$
In the given sequence,
$a_1=0, a_2=-4, a_3=-8$
$a_2-a_1=-4-(0)=-4$
$a_3-a_2=-8-(-4)=-8+4=-4$
$a_2 - a_1 = a_3 - a_2$
$d=a_2 - a_1=-4$
$a_5=a_4+d=-12+(-4)=-16$
$a_6=a_5+d=-16+(-4)=-20$
$a_7=a_6+d=-20+(-4)=-24$
(viii) Given sequence is $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, …….$
In the given sequence,
$a_1=-\frac{1}{2}, a_2=-\frac{1}{2}, a_3=-\frac{1}{2}$
$a_2-a_1=-\frac{1}{2}-(-\frac{1}{2})=-\frac{1}{2}+-\frac{1}{2}=0$
$a_3-a_2=-\frac{1}{2}-(-\frac{1}{2})=-\frac{1}{2}+-\frac{1}{2}=0$
$a_2 - a_1 = a_3 - a_2$
$d=a_2 - a_1=-\frac{1}{2}$
$a_5=a_4+d=-\frac{1}{2}+0=-\frac{1}{2}$
$a_6=a_5+d=-\frac{1}{2}+0=-\frac{1}{2}$
$a_7=a_6+d=-\frac{1}{2}+0=-\frac{1}{2}$
(ix) Given sequence is $1, 3, 9, 27, …….$
In the given sequence,
$a_1=1, a_2=3, a_3=9$
$a_2-a_1=3-1=2$
$a_3-a_2=9-3=6$
$a_2 - a_1 ≠ a_3 - a_2$
Therefore, the given sequence is not an AP.
(x) Given sequence is $a, 2a, 3a, 4a, …….$
In the given sequence,
$a_1=a, a_2=2a, a_3=3a$
$a_2-a_1=2a-a=a$
$a_3-a_2=3a-2a=a$
$a_2 - a_1 = a_3 - a_2$
$d=a_2 - a_1=a$
$a_5=a_4+d=4a+a=5a$
$a_6=a_5+d=5a+a=6a$
$a_7=a_6+d=6a+a=7a$
(xi) Given sequence is $a, a^2, a^3, a^4, …….$
In the given sequence,
$a_1=a, a_2=a^2, a_3=a^3$
$a_2-a_1=a^2-a=a(a-1)$
$a_3-a_2=a^3-a^2=a^2(a-1)$
$a_2 - a_1 ≠ a_3 - a_2$
Therefore, the given sequence is not an AP.
(xii) Given sequence is $\sqrt2, \sqrt8, \sqrt{18}, \sqrt{32}, …..$
$\sqrt8=2\sqrt2$
$\sqrt{18}=3\sqrt2$
$\sqrt{32}=4\sqrt2$
Given sequence can be written as $\sqrt2, 2\sqrt2, 3\sqrt{2}, 4\sqrt{2}, …..$
In the given sequence,
$a_1=\sqrt2, a_2=2\sqrt2, a_3=3\sqrt2$
$a_2-a_1=2\sqrt2-\sqrt2=\sqrt2$
$a_3-a_2=3\sqrt2-2\sqrt2=\sqrt2$
$a_2 - a_1 = a_3 - a_2$
$d=a_2 - a_1=\sqrt2$
$a_5=a_4+d=4\sqrt2+ \sqrt2=5\sqrt2$
$a_6=a_5+d=5\sqrt2 + \sqrt2=6\sqrt2$
$a_7=a_6+d=6\sqrt2+ \sqrt2=7\sqrt2$
(xiii) Given sequence is $\sqrt3, \sqrt6, \sqrt9, \sqrt{12}, …..$
In the given sequence,
$a_1=\sqrt3, a_2=\sqrt6, a_3=\sqrt9$
$a_2-a_1=\sqrt6-\sqrt3$
$a_3-a_2=\sqrt9-\sqrt6=3-\sqrt6$
$a_2 - a_1 ≠ a_3 - a_2$
Therefore, the given sequence is not an AP.
(xiv) Given sequence is $1^2, 3^2, 5^2, 7^2, ……$
In the given sequence,
$a_1=1^2=1, a_2=3^2=9, a_3=5^2=25$
$a_2-a_1=9-1=8$
$a_3-a_2=25-9=16$
$a_2 - a_1 ≠ a_3 - a_2$
Therefore, the given sequence is not an AP.
(xv) Given sequence is $1^2, 5^2, 7^2, 73, ……$
In the given sequence,
$a_1=1^2=1, a_2=5^2=25, a_3=7^2=49$
$a_2-a_1=25-1=24$
$a_3-a_2=49-25=24$
$a_2 - a_1 = a_3 - a_2$
$d=a_2 - a_1=24$
$a_5=a_4+d=73+24=97$
$a_6=a_5+d=97+24=121$
$a_7=a_6+d=121+24=145$
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