# Which of the following sequences are arithmetic progressions.

Question:

Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.

(i) 3, 6, 12, 24, ...

(ii) 0, −4, −8, −12, ...

(iii) $\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \ldots$

(iv) 12, 2, −8, −18, ...

(v) 3, 3, 3, 3, ...

(vi) $p, p+90, p+180 p+270, \ldots$ where $p=(999)^{989}$

(vii) $1.0,1.7,2.4,3.1, \ldots$

(viii) $-225,-425,-625,-825, \ldots$

(ix) $10,10+2^{5}, 10+2^{6}, 10+2^{7}, \ldots$

$(x) a+b,(a+1)+b,(a+1)+(b+1),(a+2)+(b+1),(a+2)+(b+2), \ldots$

(xi) $1^{2}, 3^{2}, 5^{2}, 7^{2}, \ldots$

(xii) $1^{2}, 5^{2}, 7^{2}, 73, \ldots$

Solution:

In the given problem, we are given various sequences.

We need to find out that the given sequences are an A.P or not and then find its common difference (d)

(i) $3,6,12,24, \ldots$

Here,

First term (a) = 3

$a_{1}=6$

$a_{2}=12$

Now, for the given to sequence to be an A.P,

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here,

$a_{1}-a=6-3$

$=3$

Also,

$a_{2}-a_{1}=12-6$

$=6$

Since $a_{1}-a \neq a_{2}-a_{1}$

Hence, the given sequence is not an A.P

(ii) $0,-4,-8,-12, \ldots$

Here,

First term (a) = 0

$a_{1}=-4$

$a_{2}=-8$

Now, for the given sequence to be an A.P,

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here

$a_{1}-a=-4-0$

$=-4$

Also,

$a_{2}-a_{1}=-8-(-4)$

$=-4$

Since $a_{1}-a=a_{2}-a_{1}$

Hence, the given sequence is an A.P and its common difference is $d=-4$

(iii) $\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \ldots \ldots \ldots$

Here,

First term $(a)=\frac{1}{2}$

$a_{1}=\frac{1}{4}$

$a_{2}=\frac{1}{6}$

Now, for the given to sequence to be an A.P,

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here,

$a_{1}-a=\frac{1}{4}-\frac{1}{2}$

$=\frac{1-2}{4}$

$=\frac{-1}{4}$

Also,

$a_{2}-a_{1}=\frac{1}{6}-\frac{1}{4}$

$=\frac{2-3}{12}$

$=\frac{-1}{12}$

Since $a_{1}-a \neq a_{2}-a_{1}$

Hence, the given sequence is not an A.P

(iv) $12,2,-8,-18$

Here,

First term (a) = 12

$a_{1}=2$

$a_{2}=-8$

Now, for the given to sequence to be an A.P.

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here,

$a_{1}-a=2-12$

$=-10$

Also,

$a_{2}-a_{1}=-8-2$

$=-10$

Since $a_{1}-a=a_{2}-a_{1}$

Hence, the given sequence is an A.P with the common difference $d=-10$

(v) $3,3,3,3, \ldots$

Here,

First term (a) = 3

$a_{1}=3$

$a_{2}=3$

Now, for the given to sequence to be an A.P.

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here,

$a_{1}-a=3-3$

$=0$

Also,

$a_{2}-a_{1}=3-3$

$=0$

Since $a_{1}-a=a_{2}-a_{1}$

Hence, the given sequence is an A.P and its common difference is $d=0$

(vi) $p, p+90, p+180, p+270, \ldots$ Where, $p=(999)^{99}$

Here,

First term $(a)=p$

$a_{1}=p+90$

$a_{2}=p+180$

Now, for the given to sequence to be an A.P,

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here,

$a_{1}-a=p+90-p$

$=90$

Since $a_{1}-a=a_{2}-a_{1}$

Hence, the given sequence is an A.P and its common difference is $d=90$

(vii) $1.0,1.7,2.4,3.1, \ldots .$

Here,

First term $(a)=1.0$

$a_{1}=1.7$

$a_{2}=2.4$

Now, for the given to sequence to be an A.P,

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here,

$a_{1}-a=1.7-1.0$

$=0.7$

Also,

$a_{2}-a_{1}=2.4-1.7$

$=0.7$

Since $a_{1}-a=a_{2}-a_{1}$

Hence, the given sequence is an A.P and its common difference is $d=0.7$

(viii) $-225,-425,-625,-825, \ldots \ldots$

Here,

First term $(a)=-225$

$a_{1}=-425$

$a_{2}=-625$

Now, for the given to sequence to be an A.P.

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here,

$a_{1}-a=-425-(-225)$

$=-200$

Also,

$a_{2}-a_{1}=-625-(-425)$

$=-200$

Since $a_{1}-a=a_{2}-a_{1}$

Hence, the given sequence is an A.P and its common difference is $d=-200$

(ix) $10,10+2^{5}, 10+2^{6}, 10+2^{7}, \ldots .$

Here,

First term (a) = 10

$a_{1}=10+2^{5}$

$a_{2}=10+2^{6}$

$a_{3}=10+2^{7}$

Now, for the given to sequence to be an A.P,

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here,

$a_{2}-a_{1}=10+2^{6}-10-2^{5}$

$=64-32$

$=32$

Also,

$a_{3}-a_{2}=10+2^{7}-10-2^{6}$

$=128-64$

$=64$

Since $a_{1}-a \neq a_{2}-a_{1}$

Hence, the given sequence is not an A.P

$(x) a+b,(a+1)+b,(a+1)+(b+1),(a+2)+(b+1),(a+2)+(b+2), \ldots .$

Here,

First term (a) = + b

$a_{1}=(a+1)+b$

$a_{2}=(a+1)+(b+1)$

Now, for the given to sequence to be an A.P,

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here,

$a_{1}-a=a+1+b-a-b$

$=1$

Also,

$a_{2}-a_{1}=a+1+b+1-a-1-b$

$=1$

Since $a_{1}-a=a_{2}-a_{1}$

Hence, the given sequence is an A.P and its common difference is $d=1$

(xi) $1^{2}, 3^{2}, 5^{2}, 7^{2} \ldots$

Here,

First term $(a)=1^{2}$

$a_{1}=3^{2}$

$a_{2}=5^{2}$

Now, for the given to sequence to be an A.P,

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here,

$a_{1}-a=3^{2}-1^{2}$

$=9-1$

$=8$

Also,

$a_{2}-a_{1}=5^{2}-3^{2}$

$=25-9$

$=16$

Since $a_{1}-a \neq a_{2}-a_{1}$

Hence, the given sequence is not an A.P

(xii) $1^{2}, 5^{2}, 7^{2}, 73 \ldots$

Here,

First term $(a)=1^{2}$

$a_{1}=5^{2}$

$a_{2}=7^{2}$

Now, for the given to sequence to be an A.P.

Common difference $(d)=a_{1}-a=a_{2}-a_{1}$

Here,

$a_{2}-a_{1}=5^{2}-1^{2}$

$=25-1$

$=24$

Also,

$a_{3}-a_{2}=7^{2}-5^{2}=49-25=24$

Since $a_{1}-a=a_{2}-a_{1}$

Hence, the given sequence is an A.P with the common difference $d=24$.