the 11th term of a geometric sequence is 81 and the 14th term is 3. find the 4th term
bn=b1qn−1b_n=b_1q^{n-1}bn=b1qn−1
b11=b1q10 ⟺ b1=b11q10b_{11}=b_1q^{10}\iff b_1=\frac{b_{11}}{q^{10}}b11=b1q10⟺b1=q10b11
b14=b1q13 ⟺ b14=b11q10∗q13=b11q3 ⟺ q3=b14b11b_{14}=b_1q^{13}\iff b_{14}=\frac{b_{11}}{q^{10}}*q^{13}=b_{11}q^3\iff q^3=\frac{b_{14}}{b_{11}}b14=b1q13⟺b14=q10b11∗q13=b11q3⟺q3=b11b14
q3=813=127q^3=\frac{81}{3}=\frac{1}{27}q3=381=271
q=1273=13q=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}q=3271=31
b1=811310=34∗310=314b_1=\frac{81}{\frac{1}{3}^{10}}=3^4*3^{10}=3^{14}b1=311081=34∗310=314
b4=b1q3=3143−3=311=177147b_4=b_1q^3=3^{14}3^{-3}=3^{11}=177147b4=b1q3=3143−3=311=177147
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