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Home » The     term of an A.P. is equal to three times its     term. If its     term is     , find the A.P.
The

    \[{{19}^{th}}\]

term of an A.P. is equal to three times its

    \[{{6}^{th}}\]

term. If its

    \[{{9}^{th}}\]

term is

    \[19\]

, find the A.P.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
The

    \[{{19}^{th}}\]

term of an A.P. is equal to three times its

    \[{{6}^{th}}\]

term. If its

    \[{{9}^{th}}\]

term is

    \[19\]

, find the A.P.

From the question it is given that,

    \[{{a}_{19}}~=\text{ }{{19}^{th}}~\]

term of an A.P. is equal to three times its

    \[{{6}^{th}}\]

 term =

    \[3{{a}_{6}}\]

    \[{{a}_{9}}~=\text{ }19\]

As we know, an = a + (n – 1)d

    \[{{a}_{9}}~=\text{ }a\text{ }+\text{ }8d\text{ }=\text{ }19\]

… [equation (i)]

Then,

    \[\begin{array}{*{35}{l}} ~{{a}_{19}}~=\text{ }3\left( a\text{ }+\text{ }5d \right) \\ a\text{ }+\text{ }18d\text{ }=\text{ }3a\text{ }+\text{ }15d \\ 3a\text{ }\text{ }a\text{ }=\text{ }18d\text{ }\text{ }15d \\ 2a\text{ }=\text{ }3d \\ a\text{ }=\text{ }\left( 3/2 \right)d \\ \end{array}\]

Now substitute the value of a in equation (i) we get,

    \[\begin{array}{*{35}{l}} \left( 3/2 \right)d\text{ }+\text{ }8d\text{ }=\text{ }19 \\ \left( 3d\text{ }+\text{ }16d \right)/2\text{ }=\text{ }19 \\ \left( 19/2 \right)d\text{ }=\text{ }19 \\ d\text{ }=\text{ }\left( 19\text{ }\times \text{ }2 \right)/19 \\ d\text{ }=\text{ }2 \\ \end{array}\]

To find out the value of a substitute the value of d in equation (i)

    \[\begin{array}{*{35}{l}} a\text{ }+\text{ }8d\text{ }=\text{ }19 \\ a\text{ }+\text{ }\left( 8\text{ }\times \text{ }2 \right)\text{ }=\text{ }19 \\ a\text{ }+\text{ }16\text{ }=\text{ }19 \\ a\text{ }=\text{ }19\text{ }\text{ }16 \\ a\text{ }=\text{ }3 \\ \end{array}\]

Therefore, A.P. is

    \[3,\text{ }5,\text{ }7,\text{ }9,\text{ }\ldots \]

i

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