Verify that each of the following
Question:

Verify that each of the following is an AP and then write its next three terms.

(i) $0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots$

(ii) $5, \frac{14}{3}, \frac{13}{3}, 4, \ldots$

(iii) $\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, \ldots$

(iv) $a+b,(a+1)+b,(a+1)+(b+1), \ldots$

(v) $a, 2 a+1,3 a+2,4 a+3, \ldots$

Solution:

(i) Here, $a_{1}=0, a_{2}=\frac{1}{4}, a_{3}=\frac{1}{2}$ and $a_{4}=\frac{3}{4}$

$a_{2}-a_{1}=\frac{1}{4}, a_{3}-a_{2}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}, a_{4}-a_{3}=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}$

$\therefore \quad a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}$

Since, the each successive term of the given list has the same difference. So, it forms an AP. The next three terms are, $a_{5}=a_{1}+4 d$

$=a+4\left(\frac{1}{4}\right)=1, a_{6}=a_{1}+5 d=a+5\left(\frac{1}{4}\right)=\frac{5}{4}$

$a_{7}=a+6 d=0+\frac{6}{4}=\frac{3}{2}$

(ii) Here, $a_{1}=5, a_{2}=\frac{14}{3}, a_{3}=\frac{13}{3}$ and $a_{4}=4$

$a_{2}-a_{1}=\frac{14}{3}-5=\frac{14-15}{3}=\frac{-1}{3}, a_{3}-a_{2}=\frac{13}{3}-\frac{14}{3}=-\frac{1}{3}$

$a_{4}-a_{3}=4-\frac{13}{3}=\frac{12-13}{3}=\frac{-1}{3}$

$\because \quad a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}$

Since, the each successive term of the given list has same difference.

It forms an AP.

The next three terms are,

$a_{5}=a_{1}+4 d=5+4\left(-\frac{1}{3}\right)=5-\frac{4}{3}=\frac{11}{3}$

$a_{6}=a_{1}+5 d=5+5\left(-\frac{1}{3}\right)=5-\frac{5}{3}=\frac{10}{3}$

$a_{7}=a_{1}+6 d=5+6\left(-\frac{1}{3}\right)=5-2=3$

(iii) Here, $a_{1}=\sqrt{3}, a_{2}=2 \sqrt{3}$ and $a_{3}=3 \sqrt{3}$

$a_{2}-a_{1}=2 \sqrt{3}-\sqrt{3}=\sqrt{3}, a_{3}-a_{2}=3 \sqrt{3}-2 \sqrt{3}=\sqrt{3}$

$\because \quad a_{2}-a_{1}=a_{3}-a_{2}=\sqrt{3}=$ Common difference

Since, the each successive term of the given list has same difference.

So, it forms an AP.

The next three terms are.

$a_{4}=a_{1}+3 d=\sqrt{3}+3(\sqrt{3})=4 \sqrt{3}$

$a_{5}=a_{1}+4 d=\sqrt{3}+4 \sqrt{3}=5 \sqrt{3}$

$a_{6}=a_{1}+5 d=\sqrt{3}+5 \sqrt{3}=6 \sqrt{3}$

(lv) Here, $a_{1}=a+b_{1} a_{2}=(a+1)+b_{1} a_{3}=(a+1)+(b+1)$

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

$a_{3}-a_{2}=(a+1)+(b+1)-[(a+1)+b]$

$=a+1+b+1-a-1-b=1$

$\because \quad a_{2}-a_{1}=a_{3}-a_{2}=1=$ Common difference

Since, the each successive term of the given list has same difference.

So, it forms an AP.

The next three terms are,

$a_{4}=a_{1}+3 d=a+b+3(1)=(a+2)+(b+1)$

$a_{5}=a_{1}+4 d=a+b+4(1)=(a+2)+(b+2)$

$a_{6}=a_{1}+5 d=a+b+5(1)=(a+3)+(b+2)$

(v) Here, $a_{1}=a, a_{2}=2 a+1, a_{3}=3 a+2$ and $a_{4}=4 a+3$

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

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

$a_{4}-a_{3}=4 a+3-3 a-2=a+1$

$a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}=a+1=$ Common difference

Since, the each successive term of the given list has same difference.

So, it forms an AP.

The next three terms are,

$a_{5}=a+4 d=a+4(a+1)=5 a+4$

$a_{6}=a+5 d=a+5(a+1)=6 a+5$

$a_{7}=a+6 d=a+6(a+1)=7 a+6$

 

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