Discussion
The fourth term of an A.P. is 11 and the seventh term is 20. Find the 15th term.
Answer: B
We have \(a+3d=11\) and \(a+6d=20\). Subtracting the equations gives \(3d=9\), so \(d=3\). Substituting back, \(a+3(3)=11 \Rightarrow a+9=11 \Rightarrow a=2\). The 15th term is \(a_{15} = a+14d = 2 + 14(3) = 2+42=44\).
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