Find a formula for {eq}a_n {/eq} for the arithmetic sequence.

{eq}a_5 = 190, a_{10} = 115 {/eq}

Given Data:

• The fifth and the tenth term is: {eq}{a_5} = 190,\;{\rm{and }}\;{a_{10}} = 115 {/eq}.

Let the first term is a and the common difference is d, then the nth term is given by,

{eq}{a_n} = a + \left( {n - 1} \right)d {/eq}

The given fifth term of the sequence is,

{eq}\begin{align*} {a_5} &= a + \left( {5 - 1} \right)d\\ 190 &= a + 4d \cdots \cdots \cdots (1) \end{align*} {/eq}

The given tenth term of the sequence is,

{eq}\begin{align*} {a_{10}} &= a + \left( {10 - 1} \right)d\\ 115 &= a + 9d \cdots \cdots \cdots (2) \end{align*} {/eq}

Subtract equation (2) from equation (1). We have,

{eq}\begin{align*} 190 - 115 &= a + 4d - \left( {a + 9d} \right)\\ 75 &= a + 4d - a - 9d\\ - 5d &= 75\\ d &= - 15 \end{align*} {/eq}

Substitute the value of d in equation (1).

{eq}\begin{align*} 190 &= a + 4\left( { - 15} \right)\\ 190 &= a - 60\\ a &= 190 + 60\\ a &= 250 \end{align*} {/eq}

Substitute the value of a and d in the nth term. We have,

{eq}\begin{align*} {a_n} &= a + \left( {n - 1} \right)d\\ &= 250 + \left( {n - 1} \right)\left( { - 15} \right)\\ &= 250 - 15n + 15\\ {a_n} &= 265 - 15n \end{align*} {/eq}

Thus, the required formula is {eq}{a_n} = 265 - 15n {/eq}.