The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?

Firewolf Dec 5, 2018

#1**+1 **

\(a_n = a_1 + (n-1)d\\ a_5 = a_1 + 4d = 9\\ a_{32} = a_1 + 31d = -84\\ \text{subtract }a_5 \text{ from }a_{32}\\ 27d=-93\\ d=-\dfrac{93}{27} = -\dfrac{31}{9}\\ a_1 = 9-4d = 9-4\left(-\dfrac{31}{9}\right) = \\ 9+\dfrac{124}{9} = \dfrac{205}{9}\\ a_{23} = a_1 + 22d = \dfrac{205}{9}+22\left(-\dfrac{31}{9}\right) = -53 \)

.Rom Dec 5, 2018

#2**+11 **

**The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?**

Formula of an arithmetic sequence:

\(\begin{array}{|rcll|} \hline \begin{vmatrix} a_i & a_j & a_k \\ i & j & k \\ 1 & 1 & 1 & \\ \end{vmatrix} = 0 \\\\ a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\ \hline \end{array} \)

\(\text{Set $i=5$, $\ j=32$, $\ k=23$ } \\ \text{Set $a_i=a_5=9$, $\ a_j=a_{32}=-84$, $\ a_k=a_{23}$ } \)

\(\begin{array}{|rcll|} \hline a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\\\ 9(32-23)+(-84)(23-5)+a_{23}(5-32) &=& 0 \\\\ 9(9)+(-84)(18)-a_{23}(27) &=& 0 \\\\ 81-1512-27a_{23} &=& 0 \\\\ 27a_{23} &=& 81-1512 \\\\ 27a_{23} &=& -1431 \\\\ a_{23} &=& -\dfrac{1431}{27} \\\\ \mathbf{a_{23}} &\mathbf{=}& \mathbf{-53} \\ \hline \end{array}\)

heureka Dec 6, 2018