![\[{{19}^{th}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-11343e55db4dcbd3658f45922dc0ab47_l3.png "Rendered by QuickLaTeX.com")
![\[{{9}^{th}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4b27f8dc0fc1ef5884f7984f9f2cc502_l3.png "Rendered by QuickLaTeX.com")
![\[19\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a828a599e50d3048b3a6c572e4fad3b6_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that,
![\[{{a}_{19}}~=\text{ }{{19}^{th}}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1cbf303b70b53599fee4d3518eefc21d_l3.png "Rendered by QuickLaTeX.com")
term of an A.P. is equal to three times its
![\[{{6}^{th}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2b89494afc708fa36d5cbe8dc14cfaec_l3.png "Rendered by QuickLaTeX.com")
term =
![\[3{{a}_{6}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-124682d97ec76139ed0863f018c4856d_l3.png "Rendered by QuickLaTeX.com")
![\[{{a}_{9}}~=\text{ }19\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ea357d5b03f1fbde72a1119a0f2a14b1_l3.png "Rendered by QuickLaTeX.com")
As we know, an = a + (n – 1)d
![\[{{a}_{9}}~=\text{ }a\text{ }+\text{ }8d\text{ }=\text{ }19\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-228ab692f66391afeec8d5877518225c_l3.png "Rendered by QuickLaTeX.com")
… [equation (i)]
Then,
![\[\begin{array}{*{35}{l}} ~{{a}_{19}}~=\text{ }3\left( a\text{ }+\text{ }5d \right) \\ a\text{ }+\text{ }18d\text{ }=\text{ }3a\text{ }+\text{ }15d \\ 3a\text{ }\text{ }a\text{ }=\text{ }18d\text{ }\text{ }15d \\ 2a\text{ }=\text{ }3d \\ a\text{ }=\text{ }\left( 3/2 \right)d \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e18849fee45abfe9013b7fd85f9bc696_l3.png "Rendered by QuickLaTeX.com")
Now substitute the value of a in equation (i) we get,
![\[\begin{array}{*{35}{l}} \left( 3/2 \right)d\text{ }+\text{ }8d\text{ }=\text{ }19 \\ \left( 3d\text{ }+\text{ }16d \right)/2\text{ }=\text{ }19 \\ \left( 19/2 \right)d\text{ }=\text{ }19 \\ d\text{ }=\text{ }\left( 19\text{ }\times \text{ }2 \right)/19 \\ d\text{ }=\text{ }2 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f904c73ec0afa8d13fe5eb9f67908b02_l3.png "Rendered by QuickLaTeX.com")
To find out the value of a substitute the value of d in equation (i)
![\[\begin{array}{*{35}{l}} a\text{ }+\text{ }8d\text{ }=\text{ }19 \\ a\text{ }+\text{ }\left( 8\text{ }\times \text{ }2 \right)\text{ }=\text{ }19 \\ a\text{ }+\text{ }16\text{ }=\text{ }19 \\ a\text{ }=\text{ }19\text{ }\text{ }16 \\ a\text{ }=\text{ }3 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1c4586b10c32dccc82fee3d9716c8b6f_l3.png "Rendered by QuickLaTeX.com")
Therefore, A.P. is
![\[3,\text{ }5,\text{ }7,\text{ }9,\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4230645b5bf92c9be62567c8b951abd9_l3.png "Rendered by QuickLaTeX.com")
