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# question

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327
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The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?

May 18, 2020

#1
+22158
0

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

----------------------

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

May 18, 2020
#2
+26213
+1

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

May 19, 2020