The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?
The formula for the nth term of an arithmetic sequence is: tn = t1 + (n - 1)d
where tn = nth term t1 = 1st term d = common difference
t5 = 9 ---> t5 = t1 + (5 - 1)d ---> 9 = t1 + 4d
t32 = -84 ---> t32 = t1 + (32 - 1)d ---> -84 = t1 + 31d
9 = t1 + 4d ---> 9 = t1 + 4d
-84 = t1 + 31d ---> 84 = - t1 - 31d
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93 = -27d
d = 93/-27 = -31/9
If d = -31/9 ---> 9 = t1 + 4d ---> 9 = t1 + 4(-31/9)
9 = t1 - 124/9
t1 = 205/9
tn = t1 + (n - 1)d --> t23 = 205/9 + (23 - 1)(-31/9)
= -53
The fifth term of an arithmetic sequence is 9 and the 32nd term is -84.
What is the 23rd term?
\(\begin{array}{|rcll|} \hline i < j < k \\ a_i &=& a+(i-1)d \\ a_j &=& a+(j-1)d \\ a_k &=& a+(k-1)d \\ \hline \mathbf{a_j-a_i } &=& \mathbf{(j-1)d-(i-1)d} \\ a_j-a_i&=& d\Big((j-1)-(i-1)\Big) \\ \mathbf{a_j-a_i} &=& \mathbf{ d(j-i) } \\ \\ \mathbf{a_k-a_i} &=& \mathbf{(k-1)d-(i-1)d} \\ a_k-a_i&=& d\Big((k-1)-(i-1)\Big) \\ \mathbf{a_k-a_i} &=& \mathbf{ d(k-i) } \\ \hline \dfrac{a_j-a_i}{a_k-a_i} &=& \dfrac{ d(j-i)}{d(k-i)} \\\\ \dfrac{a_j-a_i}{a_k-a_i} &=& \dfrac{ (j-i)}{(k-i)} \\ (a_j-a_i)(k-i) &=& (a_k-a_i)(j-i) \\ a_j(k-i)-a_i(k-i) &=& a_k(j-i)-a_i(j-i) \\ a_j(k-i) &=& a_k(j-i)+a_i(k-i)-a_i(j-i) \\ a_j(k-i) &=& a_k(j-i)+a_i\Big(k-i)-(j-i)\Big) \\ a_j(k-i) &=& a_k(j-i)+a_i(k-j) \\ \mathbf{a_j} &=& \mathbf{a_k\frac{j-i}{k-i}+a_i\frac{k-j}{k-i}} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{a_j} &=& \mathbf{a_k\frac{j-i}{k-i}+a_i\frac{k-j}{k-i}} \\ && \boxed{a_{5} =9=a_i\quad i = 5\\ a_{23}=a_j \quad j=23\\ a_{32}=-84=a_k\quad k=32} \\\\ a_{23} &=& (-84)*(\frac{23-5}{32-5})+ 9*(\frac{32-23}{32-5}) \\\\ a_{23} &=& (-84)*(\frac{18}{27})+ 9*(\frac{9}{27}) \\\\ a_{23} &=&(-84)*(\frac{2}{3})+ 9*(\frac{1}{3}) \\\\ a_{23} &=& -\frac{84*2}{3}+ \frac{9}{3} \\\\ a_{23} &=& - 28*2 + 3 \\\\ \mathbf{a_{23}} &=& \mathbf{-53} \\ \hline \end{array}\)
The 23rd term is -53