Find the \[\mathbf{3}{{\mathbf{1}}^{\mathbf{st}}}\] term of an A.P. whose \[\mathbf{1}{{\mathbf{1}}^{\mathbf{th}}}\] term is \[38\] and \[{{6}^{th}}\] term is \[73\].

Find the

$\mathbf{3}{{\mathbf{1}}^{\mathbf{st}}}$

term of an A.P. whose

$\mathbf{1}{{\mathbf{1}}^{\mathbf{th}}}$

term is

$38$

and

${{6}^{th}}$

term is

$73$

Find the

$\mathbf{3}{{\mathbf{1}}^{\mathbf{st}}}$

term of an A.P. whose

$\mathbf{1}{{\mathbf{1}}^{\mathbf{th}}}$

term is

$38$

and

${{6}^{th}}$

term is

$73$

From the question it is given that,

$\begin{array}{*{35}{l}} {{T}_{11}}~=\text{ }38 \\ {{T}_{6}}~=\text{ }73 \\ \end{array}$

Let us assume ‘a’ be the first term and ‘d’ be the common difference,

So,

${{T}_{11}}~=\text{ }a\text{ }+\text{ }10d\text{ }=\text{ }38$

equation (i)

${{T}_{11}}~=\text{ }a\text{ }+\text{ }10d\text{ }=\text{ }38$

equation (ii)

Subtracting both equation (i) and equation (i),

$\begin{array}{*{35}{l}} \left( a\text{ }+\text{ }10d \right)\text{ }\text{ }\left( a\text{ }+\text{ }5d \right)\text{ }=\text{ }73\text{ }\text{ }38 \\ a\text{ }+\text{ }10d\text{ }\text{ }a\text{ }\text{ }5d\text{ }=\text{ }35 \\ 5d\text{ }=\text{ }35 \\ d\text{ }=\text{ }35/5 \\ d\text{ }=\text{ }7 \\ \end{array}$

now, substitute the value of d in equation (i) to find out a, we get

$\begin{array}{*{35}{l}} a\text{ }+\text{ }10d\text{ }=\text{ }38 \\ a\text{ }+\text{ }10\left( 7 \right)\text{ }=\text{ }38 \\ a\text{ }+\text{ }70\text{ }=\text{ }38 \\ a\text{ }=\text{ }38\text{ }\text{ }70 \\ a\text{ }=\text{ }-32 \\ \end{array}$

Therefore,

$\begin{array}{*{35}{l}} {{T}_{31}}~=\text{ }a\text{ }+\text{ }30d \\ =\text{ }-32\text{ }+\text{ }30\left( 7 \right) \\ =\text{ }-32\text{ }+\text{ }210 \\ =\text{ }178 \\ \end{array}$

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