Geometric sequence 5

About members of geometric sequence we know:

3a5:a3=27:25 3 a5:a3 = 27:25

7a3+5a7=1:564 7 a3 +5 a7 = 1 : 564


Calculate a1 (first member) and q (common ratio or q-coefficient)


Result

q =  0.6
a1 =  204.84658

Solution:

3a5:a3=27:25 3a1q4:(a1q2)=27:25 3q2=27/25 q2=9/25 q=±3/5 q1=3/5 q2=3/5  3 a_5:a_3 = 27:25 \ \\ 3 a_1 q^4 : (a_1 q^2 ) = 27:25 \ \\ 3 q^2 = 27/25 \ \\ q^2 = 9/25 \ \\ q = \pm 3/5 \ \\ q_1 = - 3/5 \ \\ q_2 = 3/5 \ \\ \ \\
7a3+5a7=1:564 7a1q2+5a1q6=1:564 a1(7q2+5q6)=1:564  a1=564:(63/25+729/3125) a1=204.846587 a_3 +5 a_7 = 1 : 564 \ \\ 7a_1q^2+5a_1q^6 = 1:564 \ \\ a_1 (7q^2+5q^6) = 1:564 \ \\ \ \\ a_1 = 564 : (63/25 + 729/3125) \ \\ a_1= 204.84658



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