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Find the common ratio of the geometric sequence {a_(n)}_(n=1)^(oo) given that: {:[a_(2)=-(1)/(4)],[,a_(6)=-(12)/(2//3)]:}

Find the common ratio of the geometric sequence {an}n=1∞ \left\{a_{n}\right\}_{n=1}^{\infty} given that:\newlinea2=−14a6=−122/3\begin{array}{ll} a_{2}=-\frac{1}{4} \\ a_{6}=-\frac{12}{2 / 3}\end{array}

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Full solution

Q. Find the common ratio of the geometric sequence {an}n=1∞ \left\{a_{n}\right\}_{n=1}^{\infty} given that:\newlinea2=−14a6=−122/3\begin{array}{ll} a_{2}=-\frac{1}{4} \\ a_{6}=-\frac{12}{2 / 3}\end{array}
  1. Identify first term: Identify the first term (a1a_1) of the geometric sequence.\newlineGiven that a1=−14a_1 = -\frac{1}{4}, we can use this value to find the common ratio.
  2. Identify sixth term: Identify the sixth term a6a_6 of the geometric sequence.\newlineGiven that a6=−1223a_6 = -\frac{12}{\frac{2}{3}}, we need to simplify this fraction to find the actual value of a6a_6.\newlinea6=−12×(32)=−18a_6 = -12 \times \left(\frac{3}{2}\right) = -18
  3. Find common ratio: Use the formula for the nth term of a geometric sequence to find the common ratio rr. The nth term of a geometric sequence is given by an=a1⋅r(n−1)a_n = a_1 \cdot r^{(n-1)}. We can use the first term a1a_1 and the sixth term a6a_6 to find the common ratio rr. a6=a1⋅r(6−1)a_6 = a_1 \cdot r^{(6-1)} −18=(−14)⋅r5-18 = (-\frac{1}{4}) \cdot r^5
  4. Solve for ratio: Solve for the common ratio rr. To isolate rr, we divide both sides of the equation by a1a_1. r5=−18/(−1/4)r^5 = -18 / (-1/4) r5=18×(4/1)r^5 = 18 \times (4/1) r5=72r^5 = 72 Now, take the fifth root of both sides to solve for rr. r=721/5r = 72^{1/5}
  5. Calculate value of r: Calculate the fifth root of 7272 to find the value of rr.\newliner≈2.6390158215457884r \approx 2.6390158215457884 (using a calculator)

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