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 in an arithmatic series x, then 7 then y, then z, their sum equals 12, what is the first term and common difference

 Dec 12, 2016
 #1
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Your AS is: First term=15, common difference = -8

15, 7, -1, -9

 Dec 12, 2016
 #2
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in an arithmatic series x, then 7 then y, then z, their sum equals 12,

what is the first term and common difference

 

\(\begin{array}{|lrcll|} \hline & x = a_1 &=& a_1 + 0\cdot d \\ & 7 = a_2 &=& a_1 + 1\cdot d \\ & y = a_3 &=& a_1 + 2\cdot d \\ & z = a_4 &=& a_1 + 3\cdot d \\\\ (1) & 7 &=& a_1 + d \\ & a_1 &=& 7 - d \\\\ (2)& 12 &=& 4 a_1 + 6d \\ & 12 &=& 4(7-d) + 6d \\ & 12 &=& 28-4d + 6d \\ & 12 &=& 28+2d \quad | \quad : 2\\ & 6 &=& 14 +d \\ & d &=& 6-14\\ & \mathbf{d} & \mathbf{=} & \mathbf{-8} \\\\ & a_1 &=& 7 - d \\ & a_1 &=& 7 - (-8) \\ & a_1 &=& 7+8 \\ & \mathbf{a_1} & \mathbf{=} & \mathbf{15} \\ \hline \end{array} \)

 

The first term is 15 and common difference is -8

\(15,\ 7,\ -1,\ -9,\)


\(x= 15 \\ y=-1 \\ z=-9 \\\)

 

laugh

 Dec 12, 2016
edited by heureka  Dec 12, 2016

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