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Sagot :
Given that for an Arithmetic progression:
[tex]\begin{gathered} \text{ The third term T}_3=94 \\ \text{ The sixth term T}_6=85 \end{gathered}[/tex]The n-th term of the sequence is given by the formula:
[tex]\begin{gathered} U_n=U_1+(n-1)d \\ \text{where }U_1=\text{ first tem} \\ d\text{ = common difference} \end{gathered}[/tex]Hence,
[tex]\begin{gathered} T_3=U_1+(3-1)d=94 \\ U_1+2d=94\ldots\ldots(1) \\ T_6=U_1+(6-1)d\text{ = 85} \\ T_6=U_1+5d\text{ = 85}\ldots\ldots\text{.}(2) \end{gathered}[/tex]Solving both equations simultaneously using the elimination method,
[tex]\begin{gathered} U_1+2d=94\ldots\ldots(1) \\ U_1+5d\text{ = 85}\ldots\ldots\text{.}(2) \end{gathered}[/tex]Subtract equation (2) from (1)
[tex]\begin{gathered} 2d-5d=94-85 \\ -3d=9 \\ d=\frac{9}{-3} \\ d=-3 \end{gathered}[/tex]To find the first term, substitute d = -3 into equation (1)
[tex]\begin{gathered} U_1+2d=94 \\ U_1+2(-3)=94 \\ U_1-6=94 \\ U_1=94+6 \\ U_1=100 \end{gathered}[/tex]To find the 10th term of the sequence.
[tex]\begin{gathered} T_{10}=a+(10-1)d \\ T_{10}=a+9d \\ T_{10}=100+9(-3) \\ T_{10}=100-27 \\ T_{10}=73 \end{gathered}[/tex]Therefore, the 10th term of the arithmetic sequence is 73
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