Find the A.P. whose nth term is \[\mathbf{7}\text{ }\text{ }\mathbf{3K}\]. Also find the \[20\]th term.

Find the A.P. whose nth term is

$\mathbf{7}\text{ }\text{ }\mathbf{3K}$

. Also find the

$20$

th term.

Find the A.P. whose nth term is

$\mathbf{7}\text{ }\text{ }\mathbf{3K}$

. Also find the

$20$

th term.

From the question it is given that,

n^th term is

$\mathbf{7}\text{ }\text{ }\mathbf{3K}$

So,

${{T}_{n}}~=\text{ }7\text{ }\text{ }3n$

Now, we start giving values,

$1,\text{ }2,\text{ }3,\text{ }\ldots$

in the place of n, we get,

$\begin{array}{*{35}{l}} {{T}_{1}}~=\text{ }7\text{ }\text{ }\left( 3\text{ }\times \text{ }1 \right)\text{ }=\text{ }7\text{ }\text{ }3\text{ }=\text{ }4 \\ {{T}_{2}}~=\text{ }7\text{ }\text{ }\left( 3\text{ }\times \text{ }2 \right)\text{ }=\text{ }7\text{ }\text{ }6\text{ }=\text{ }1 \\ {{T}_{3}}~=\text{ }7\text{ }\text{ }\left( 3\text{ }\times \text{ }3 \right)\text{ }=\text{ }7\text{ }\text{ }9\text{ }=\text{ }-2 \\ {{T}_{4}}~=\text{ }7\text{ }\text{ }\left( 3\text{ }\times \text{ }4 \right)\text{ }=\text{ }7\text{ }\text{ }12\text{ }=\text{ }\text{ }5 \\ {{T}_{20}}~=\text{ }7\text{ }\text{ }\left( 3\text{ }\times \text{ }20 \right)\text{ }=\text{ }7\text{ }\text{ }60\text{ }=\text{ }\text{ }53 \\ \end{array}$

Therefore, A.P. is

$4,\text{ }1,\text{ }-2,\text{ }-5,\text{ }\ldots$

So,

$20$

^th term is

$-53$

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