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1. Write the first terms of each of the following sequences whose {{n}^{th}} term are: (v) {{a}_{n}}={{\left( -1 \right)}^{n}}{{.2}^{n}} (vi) {{a}_{n}}=n\left( n-2 \right)/2
Arithmetic Progression | Class 10 | ICSE | Maths | RD Sharma
1. Write the first terms of each of the following sequences whose {{n}^{th}} term are: (v) {{a}_{n}}={{\left( -1 \right)}^{n}}{{.2}^{n}} (vi) {{a}_{n}}=n\left( n-2 \right)/2

An arithmetic progressions or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solutions:

(v) {{a}_{n}}={{\left( -1 \right)}^{n}}{{2}^{n}}

Given sequence whose {{a}_{n}}={{\left( -1 \right)}^{n}}{{2}^{n}}

To get first five terms of the sequence, put n=1,2,3,4,5 in the above.

{{a}_{1}}={{\left( -1 \right)}^{1}}{{.2}^{1}}=\left( -1 \right).2=-2

{{a}_{2}}={{\left( -1 \right)}^{2}}{{.2}^{2}}=\left( -1 \right).4=4

{{a}_{3}}={{\left( -1 \right)}^{3}}{{.2}^{3}}=\left( -1 \right).8=-8

{{a}_{4}}={{\left( -1 \right)}^{4}}{{.2}^{4}}=\left( -1 \right).16=16

{{a}_{5}}={{\left( -1 \right)}^{5}}{{.2}^{5}}=\left( -1 \right).32=-32

\therefore the first five terms of the sequence are -2,4,-8,16,-32.

(vi) {{a}_{n}}=n\left( n-2 \right)/2

The given sequence is,
{{a}_{n}}=\frac{n\left( n-2 \right)}{2}

To get the first five terms of the sequence, put n=1,2,3,4,5.

And, we get

{{a}_{1}}=\frac{1\left( 1-2 \right)}{2}=\frac{1-1}{2}=\frac{-1}{2}

{{a}_{2}}=\frac{2\left( 2-2 \right)}{2}=\frac{2.0}{2}=0

{{a}_{3}}=\frac{3\left( 3-2 \right)}{2}=\frac{3.1}{2}=\frac{3}{2}

{{a}_{4}}=\frac{4\left( 4-2 \right)}{2}=\frac{4.2}{2}=4

{{a}_{5}}=\frac{5\left( 5-2 \right)}{2}=\frac{5.3}{2}=\frac{15}{2}

\therefore the required first five terms are -1/2,0,3/2,4,15/2

i

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