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i)

Write the first three terms of the APs when a and d are as given below:

\mathrm{a}=\frac{1}{2}, \mathrm{~d}=-\frac{1}{6}

ii)

Write the first three terms of the APs when a and d are as given below:

a = – 5, d = – 3

iii)

Write the first three terms of the APs when a and d are as given below:

\mathrm{a}=\sqrt{2}, \mathrm{~d}=\frac{1}{\sqrt{2}}

Answers (1)

i)

\\$Here a =$ \frac{1}{2} , d = -\frac{1}{6} \\ {{a}_{1}}=a=\frac{1}{2} \\ {{a}_{2}}=a+d (Using a\textsubscript{n} = a + (n-1).d)\\ a\textsubscript{3} = a + 2d\\ \begin{aligned} & =\frac{1}{2}+2\left( \frac{-1}{6} \right) \ & =\frac{1}{2}-\frac{2}{6} \ \end{aligned} \\ =\frac{3-2}{6}=\frac{1}{6} \\ $Hence, first there terms are:$ \frac{1}{2},\frac{1}{3},\frac{1}{6},..... \\

ii)

\text{Here } a = -5, \quad d = -3 \\ a_{1} = a = -5 \\ a_{2} = a + d \quad (\text{Using } a_{n} = a + (n-1)d) \\ = -5 - 3 = -8 \\ a_{3} = a + 2d \\ = -5 + 2(-3) \\ = -5 - 6 = -11 \\ \text{Here, the first three terms are: } -5, -8, -11.

iii)

\\$Here a$ = \sqrt{2},d=\frac{1}{\sqrt{2}} \\ a\textsubscript{2}=a+d (using a\textsubscript{n} = a + (n-1)d)\\ =\sqrt{2}+\frac{1}{\sqrt{2}} \\ =\frac{2+1}{\sqrt{2}}=\frac{3}{\sqrt{2}} \\ a\textsubscript{3} = a + 2d\\ =\sqrt{2}+2\left( \frac{1}{\sqrt{2}} \right) \\ =\sqrt{2}+\frac{2}{\sqrt{2}} \\ \begin{aligned} & =\frac{2+2}{\sqrt{2}} \ & =\frac{4}{\sqrt{2}} \ \end{aligned} \\ $Here, first three terms are :$ \sqrt{2},\frac{3}{\sqrt{2}},\frac{4}{\sqrt{2}} \\

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