Question

Show that each of the following sequences is an A.P. Also, find the common difference and write 3 more terms in each case.

(i) $3,-1,-5,-9\dots \dots$

(ii) $-1,\frac{1}{4},\frac{3}{2},\frac{11}{4},\dots \dots$

(iii) $\sqrt{2},3\sqrt{2},5\sqrt{2},7\sqrt{2},\dots \dots$

(iv) $9,7,5,3,\dots \dots$

Answer 1

(i) $3,-1,-5,-9\dots \dots$

$a_1=3,a_2=-1,a_3=-5,a_4=-9$

$a_2-a_1=-1-3=-4$

$a_3-a_2=-5-(-1)=-4$

$a_4-a_3=-9(-5)=-4$

$\therefore$ Common difference is d = -4

$a_4-a_3=a_3-a_2=a$

$\therefore$ The given sequence is a A.P.

$\therefore a_5=3+(5-1)(-4)=-13$

$a_6=3+(6-1)(-4)=-17$

$a_7=3+(7-1)(-4)=-21$

(ii) $-1,\frac{1}{4},\frac{3}{2},\frac{11}{4},\dots \dots$

$a_1=-1,a_2=\frac{1}{4},a_3=\frac{3}{2},a_4=\frac{11}{4}$

$a_4-a_3=a_3-a_2=a_2-a_1=\frac{5}{4}$

$\therefore$ Common difference is $d=\frac{5}{4}$

$\therefore$ The given sequence is A.P.

$a_5=-1+(5-1)\frac{5}{4}=4$

$a_6=-1+(6-1)\frac{5}{4}=\frac{21}{4}$

$a_7=-1+(7-1)\frac{5}{4}=\frac{26}{4}=\frac{13}{2}$

(iii) $\sqrt{2},3\sqrt{2},5\sqrt{2},7\sqrt{2}\dots \dots$

$a_1=\sqrt{2},a_2=3\sqrt{2},a_3=5\sqrt{2},a_4=7\sqrt{2}$

$\Rightarrow a_4-a_3=a_3-a_2=a_2-a_1=2\sqrt{2}$

$\therefore$ The common difference is $2\sqrt{2}$ and the given sequence is A.P.

$a_5=\sqrt{2}+2\sqrt{2}(5-1)=9\sqrt{2}$

$a_6=\sqrt{2}+2\sqrt{2}(6-1)=11\sqrt{2}$

$a_7=\sqrt{2}+2\sqrt{2}(7-1)=13\sqrt{2}$

(iv) $9,7,5,3,\dots \dots$

$a_1=9,a_2=7,a_3=5,a_4=3$

$a_4-a_3=a_3-a_2=a_2-a_1=-2$

$\therefore$ The common difference is -2 and the given sequence is A.P.

$a_5=9+(-2)(5-1)=1$

$a_6=9+(-2)(6-1)=-1$

$a_7=9+(-2)(7-1)=-3$