Question

calcule o valor de b de modo que a sequencia (b+1,b-3,b-5) seja uma pg

Answer 1

$\boxed{(a_2)^2=a_1*a_3}$

$(b-3)^2=(b+1)(b-5)\\\\ b^2-6b+9=b^2-5b+b-5\\\\ -6b+4b=-5-9\\\\ -2b=-14\\\\ b=\frac{14}{2}\\\\ \boxed{b=7}$