For which (r) will (r-1, 2r+2, 4r+7) be in an arithmetic progression, and what will be the common difference?

Correct answer: \(r=-2,\ d=1\)

In an arithmetic progression, the differences between consecutive terms must be equal. Here, the difference between the second and first terms is \((2r+2)-(r-1)=r+3\), while that between the third and second terms is \((4r+7)-(2r+2)=2r+5\). Thus, \(r+3=2r+5\), giving \(r=-2\). The terms then become \(-3,-2,-1\), so the common difference is \(d=1\). For \(r=0\), the two differences are 3 and 5, so that option is not correct. Exam tip: for three AP terms, also check whether \(2\times\) the middle term equals the sum of the first and third terms.