Question

Срочно Помогите пожалуйста (фото)

Answer 1

$\displaystyle\bf\\a_{1} =b_{1} +7\\\\a_{2} =b_{2} +13=b_{1} q+13\\\\a_{3} =b_{3} +15=b_{1} q^{2} +15\\\\a_{4} =b_{4} +5=b_{1} q^{3} +5\\\\\\\left \{ {{a_{2} - a_{1} =a_{4} - a_{3} } \atop {a_{2} - a_{1} = a_{3} - a_{2} }} \right. \\\\\\\left \{ {{b_{1}q+13-b_{1} -7=b_{1}q^{3}+5-b_{1} q^{2}-15 } \atop {b_{1} q+13-b_{1} -7=b_{1}q^{2} +15-b_{1} q-13 }} \right.$

$\displaystyle\bf\\:\left \{ {{b_{1} q^{3} -b_{1}q^{2}-b_{1}q+b_{1}=16 } \atop {b_{1} q^{2}-2b_{1}q+b_{1}=4}} \right. \\\\\\\frac{b_{1}(q^{3}- q^{2}-q+1) }{b_{1}(q^{2} -2q+1) } =\frac{16}{4} \\\\\\\frac{(q^{3} - q^{2})-(q-1) }{(q-1)^{2} } =4\\\\\\\frac{q^{2}(q-1)-(q-1) }{(q-1)^{2} } =4\\\\\\\frac{(q-1)(q^{2} -1)}{(q-1)^{2} } =4$

$\displaystyle\bf\\\frac{(q-1)(q-1)(q+1)}{(q-1)^{2} } =4\\\\\\q+1=4\\\\q=3\\\\b_{1} =\frac{4}{q^{2} -2q+1} =\frac{4}{3^{2} -2\cdot 3+1}=\frac{4}{4} =1\\\\b_{1} =1\\\\b_{2}=b_{1} \cdot q=1\cdot 3=3\\\\b_{3} =b_{2} \cdot q=3\cdot 3=9\\\\b_{4}=b_{3} \cdot q=9\cdot 3=27$