The sum of the first twelve terms of an Arithmetic Progression is 168. If the third term is 7, find the values of the common difference and the first term.
Question 1 Report
The sum of the first twelve terms of an Arithmetic Progression is 168. If the third term is 7, find the values of the common difference and the first term.
Answer Details
Let the first term be \(a\) and common difference \(d\).
Sum of first 12 terms is 168:
\[S_{12}=\frac{12}{2}\big(2a+11d\big)=168\Rightarrow6(2a+11d)=168\Rightarrow2a+11d=28\quad(1)\]
Third term is 7:
\[a+2d=7\quad(2)\]
From (2), \(a=7-2d\). Substitute into (1):
\[2(7-2d)+11d=28\Rightarrow14+7d=28\Rightarrow7d=14\Rightarrow d=2\]
Then \(a=7-2(2)=3\).
\[a=3,\qquad d=2\]
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