In an AP, (a_1+a_4=38) and (a_7=44). What is (a_{16})?

Correct answer: 94

Let the first term be \(a\) and the common difference be \(d\). Then \(a_1+a_4=a+(a+3d)=2a+3d=38\), while \(a_7=a+6d=44\). Multiplying the second equation by 2 gives \(2a+12d=88\). Subtracting the first equation gives \(9d=50\), so \(d=\frac{50}{9}\). Hence, \(a_{16}=a_7+9d=44+9\times\frac{50}{9}=94\). The value 80 results from incorrectly taking \(d=4\). Exam tip: eliminate the common term by multiplying one AP equation suitably before subtracting.