Login/Sign Up
Home » Show that the list of numbers     form an A.P. Find its     term and the     .
Show that the list of numbers

    \[\mathbf{9},\text{ }\mathbf{12},\text{ }\mathbf{15},\text{ }\mathbf{18},\text{ }\ldots \]

form an A.P. Find its

    \[{{16}^{th}}\]

term and the

    \[{{n}^{th}}\]

.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
Show that the list of numbers

    \[\mathbf{9},\text{ }\mathbf{12},\text{ }\mathbf{15},\text{ }\mathbf{18},\text{ }\ldots \]

form an A.P. Find its

    \[{{16}^{th}}\]

term and the

    \[{{n}^{th}}\]

.

From the question,

The first term a =

    \[9\]

Then, difference d =

    \[12\text{ }\text{ }9\text{ }=\text{ }3\]

    \[\begin{array}{*{35}{l}} 15\text{ }\text{ }12\text{ }=\text{ }3 \\ 18\text{ }\text{ }15\text{ }=\text{ }3 \\ \end{array}\]

Therefore, common difference d =

    \[3\]

From the formula,

    \[{{a}_{n}}\]

= a + (n – 1)d

    \[\begin{array}{*{35}{l}} {{T}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d \\ =\text{ }9\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)3 \\ =\text{ }9\text{ }+\text{ }3n\text{ }\text{ }3 \\ =\text{ }6\text{ }+\text{ }3n \\ \end{array}\]

So,

    \[{{T}_{16}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d\]

    \[\begin{array}{*{35}{l}}</strong> <strong> =\text{ }9\text{ }+\text{ }\left( 16\text{ }\text{ }1 \right)3 \\</strong> <strong> =\text{ }9\text{ }+\text{ }\left( 15 \right)\left( 3 \right) \\</strong> <strong> =\text{ }9\text{ }+\text{ }45 \\</strong> <strong> =\text{ }54 \\</strong> <strong>\end{array}\]

i

More Material