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Which of the following lists of numbers form an A.P.? If they form an A.P., find the common difference d and write the next three terms: (v) \[\text{ }\mathbf{10},\text{ }-\mathbf{6},\text{ }-\mathbf{2},\text{ }\mathbf{2},\text{ }\ldots \] (vi) \[{{\mathbf{1}}^{\mathbf{2}}},\text{ }{{\mathbf{3}}^{\mathbf{2}}},\text{ }{{\mathbf{5}}^{\mathbf{2}}},\text{ }{{\mathbf{7}}^{\mathbf{2}}},\text{ }\ldots \]
Which of the following lists of numbers form an A.P.? If they form an A.P., find the common difference d and write the next three terms: (v) \[\text{ }\mathbf{10},\text{ }-\mathbf{6},\text{ }-\mathbf{2},\text{ }\mathbf{2},\text{ }\ldots \] (vi) \[{{\mathbf{1}}^{\mathbf{2}}},\text{ }{{\mathbf{3}}^{\mathbf{2}}},\text{ }{{\mathbf{5}}^{\mathbf{2}}},\text{ }{{\mathbf{7}}^{\mathbf{2}}},\text{ }\ldots \]
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
Which of the following lists of numbers form an A.P.? If they form an A.P., find the common difference d and write the next three terms: (v) \[\text{ }\mathbf{10},\text{ }-\mathbf{6},\text{ }-\mathbf{2},\text{ }\mathbf{2},\text{ }\ldots \] (vi) \[{{\mathbf{1}}^{\mathbf{2}}},\text{ }{{\mathbf{3}}^{\mathbf{2}}},\text{ }{{\mathbf{5}}^{\mathbf{2}}},\text{ }{{\mathbf{7}}^{\mathbf{2}}},\text{ }\ldots \]

From the question it is given that,

First term a = \[-10\]

Then, difference d = \[-6\text{ }\text{ }\left( -\text{ }10 \right)\text{ }=\text{ }\text{ }6\text{ }+\text{ }10\text{ }=\text{ }4\]

\[\begin{array}{*{35}{l}}

-2\text{ }\text{ }\left( -6 \right)\text{ }=\text{ }\text{ }2\text{ }+\text{ }6\text{ }=\text{ }4  \\

2\text{ }\text{ }\left( -2 \right)\text{ }=\text{ }2\text{ }+\text{ }2\text{ }=\text{ }4  \\

\end{array}\]

Therefore, common difference d = \[4\]

Hence, the numbers are form A.P.

(vi) \[{{\mathbf{1}}^{\mathbf{2}}},\text{ }{{\mathbf{3}}^{\mathbf{2}}},\text{ }{{\mathbf{5}}^{\mathbf{2}}},\text{ }{{\mathbf{7}}^{\mathbf{2}}},\text{ }\ldots \]

Solution:-

From the question it is given that,

First term a \[=\text{ }{{1}^{2}}~=\text{ }1\]

Then, difference d = \[{{3}^{2}}~\text{ }{{1}^{2}}~=\text{ }9\text{ }\text{ }1\text{ }=\text{ }8\]

\[\begin{array}{*{35}{l}}

{{5}^{2}}~\text{ }{{3}^{2}}~=\text{ }25\text{ }\text{ }9\text{ }=\text{ }16  \\

{{7}^{2}}~\text{ }{{5}^{2}}~=\text{ }49\text{ }\text{ }25\text{ }=\text{ }24  \\

\end{array}\]

Therefore, common difference d is not same in the given numbers.

Hence, the numbers are not form A.P.

i

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