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  • Verify that each of the following is an AP, and then write its next three terms.0,1/4,1/2,3/4.......

i)

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

  0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots

ii)

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

5, \frac{14}{3}, \frac{13}{3}, 4, \ldots \ldots

iii)

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

\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, \ldots

iv)

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

a + b, (a + 1) + b, (a + 1) + (b + 1), ...

v)

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

a, 2a + 1, 3a + 2, 4a + 3,...

Answers (1)

i)

Here the given series is, 

 0,\frac{1}{4},\frac{1}{2},\frac{3}{4}

\\ a\textsubscript{1} = 0, a\textsubscript{2}= \frac{1}{4} {{a}_{3}} = \frac{3}{4} \\ a\textsubscript{2} - a\textsubscript{1} = \frac{1}{4}-0=\frac{1}{4} \\

\\a_3-a_2=\frac{1}{4}\\a_4-a_3=\frac{1}{4}

Here common difference is same. Hence given series is an AP 

\\a\textsubscript{5} = a\textsubscript{1}+4d $(using a\textsubscript{n} $= a\textsubscript{1 }+ (n-1).d)\\ \vspace{\baselineskip} a\textsubscript{6} = a\textsubscript{1}+ 5d \\ \vspace{\baselineskip} a\textsubscript{7} = a\textsubscript{1}+ 6d \\
\begin{aligned} & =0+6\left( \frac{1}{4} \right) \\ & =\frac{3}{2} \ \end{aligned} \\

Next three terms are

1,\frac{5}{4},\frac{3}{2} \\

ii)

\\$Here the given series is, 5, $ \frac{14}{3},\frac{13}{3},4,..... \\ $Here common difference is same. \\ Hence the given series is an AP\\$ a\textsubscript{5} = {{a}_{1}} +4d (using a\textsubscript{n} = a\textsubscript{1} + (n-1).d)\\ a\textsubscript{6 }= {{a}_{1}} +5d \\ a\textsubscript{7} = {{a}_{1}} + 6d\\ \vspace{\baselineskip} $Next terms are$ \frac{11}{3} , \frac{10}{3} ,3 \\

iii)

\\$Here the given series is, \sqrt{3},2\sqrt{3},3\sqrt{3},.... \\ $ $Here common difference is same. Hence the given series is an AP.\\$ a\textsubscript{4} = {{a}_{1}} +3d (using a\textsubscript{n} = {{a}_{1}} + (n-1).d)\\ =\sqrt{3}+3\sqrt{3}=4\sqrt{3} \\ a\textsubscript{5} = {{a}_{1}} + 4d\\ = \sqrt{3}+4\sqrt{3}=5\sqrt{3} \\ a\textsubscript{6} = {{a}_{1}} + 5d \\ = \sqrt{3}+5\sqrt{3}=6\sqrt{3} \\ $Next terms are $ 4\sqrt{3},5\sqrt{3},6\sqrt{3} \\

iv)

\\$Here the given series is, a + b, (a + 1) + b, (a + 1) + (b + 1) $ \ldots $ .\\ a\textsubscript{1} = a + b, a\textsubscript{2}= (a + 1) + b, a\textsubscript{3}= (a + 1) + (b + 1)\\ a\textsubscript{2} - a\textsubscript{1} = (a + 1) + b - a - b \\ = 1\\ a\textsubscript{3} - a\textsubscript{2} = a + 1 + b + 1 - a - 1 - b\\ = 1 \\ Here common difference is same hence given series is an AP\\ a\textsubscript{4} = a\textsubscript{1}+3d (using a\textsubscript{n} = a\textsubscript{1} + (n-1).d) \\ =(a + b) + 3(1)\\ (a + 1) + (b + 1) + 1\\ a\textsubscript{5} = a + 4d\\ =(a + b) + 4(1)\\ (a + 1) + (b + 1) + 1 + 1\\ a\textsubscript{6} = a + 5d\\ = (a + b) + 5(1)\\ (a + 1) + (b + 1) + 1 + 1 + 1\\ Next three terms are:\\ (a + 1) + (b + 1) + 1, (a + 1) + (b + 1) + 1 + 1, (a + 1) + (b + 1) + 1 + 1 + 1.\\

v)

\\$Here the given series is, a, 2a + 1, 3a + 2, 4a + 3$ \ldots $ $ \ldots $ \\ a\textsubscript{1} = a , a\textsubscript{2}=2a+1 , a\textsubscript{3}=3a + 2\\ a\textsubscript{2} - a\textsubscript{1}= 2a + 1 - a \\ = a + 1\\ a\textsubscript{3} - a\textsubscript{2} = 3a + 2 - (2a + 1)\\ =a + 1\\ Here common difference is same hence given series is an AP.\\ a\textsubscript{5} = a\textsubscript{1}+4d (using a\textsubscript{n} = a\textsubscript{1}+ (n-1)d) \\ =a + 4(a + 1)\\ = a + 4a + 4\\ = 5a + 4\\ a\textsubscript{6} = a\textsubscript{1} + 5d \\ = a + 5(a + 1)\\ = a + 5a + 5\\ = 6a + 5\\ a\textsubscript{7} = a\textsubscript{1} + 6d \\ = a + 6(a + 1)\\ = a + 6a + 6\\ = 7a + 6\\ Next three terms are: 5a + 4, 6a + 5, 7a + 6 $ \ldots $ \\

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