Find a formula for {eq}a_n {/eq} for the arithmetic sequence.
{eq}a_5 = 190, a_{10} = 115 {/eq}
Question:
Find a formula for {eq}a_n {/eq} for the arithmetic sequence.
{eq}a_5 = 190, a_{10} = 115 {/eq}
Arithmetic Sequence and Common Difference:
In mathematics, an arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. If the first and the second terms are {eq}a,\;{\rm{and}}\;b {/eq}, then the common difference of the arithmetic sequence is {eq}d = b - a {/eq}. The nth term of an arithmetic sequence is given by, {eq}{a_n} = a + \left( {n - 1} \right)d {/eq}.
Answer and Explanation: 1
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}.
Learn more about this topic:
from
Chapter 21 / Lesson 4Learn the Arithmetic sequence formula and meaning. Discover how to find the common difference and read arithmetic sequence examples.