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(2010 Susong County three modules) arithmetic {an} series of the former n items and Sn, S9=-18, S13=-52, geometric ratio {bn}, b5=a5, b7=a7, the value of B15 is () A. 64B. -64C. 128D. -128
∵ S9 = 92 (a1 + A9) = 9A5 = - 18, S13 = 132 (a1 + A13) = 13a7 = - 52, ∵ A5 = - 2, a7 = - 4, and ∵ B5 = A5, B7 = A7, ∵ B5 = - 2, B7 = - 4, ∵ Q2 = 2, B15 = B7 · Q8 = - 4 × 16 = - 64
In the arithmetic sequence {an}, S9 = - 18, S13 = = - 52, in the arithmetic sequence {BN}, B5 = A5, B7 = A7, then the geometric mean of B19, B1?
Answer 8 * under the root 2
Newspaper 13-5
Solution, arithmetic sequence
S9 = 9(a1 + a9 )/2 =9a5 =-18
S13 = 13(a1+a13)/2 =13a7=-52
So, A5 = - 2,
a7=-4
If B5 = A5, B7 = A7 in {BN}, then
q^2=b7/b5=a7/a5=2
B5 = b1q ^ 4 = - 4b1 = - 2, then B1 = - 1 / 2
B19, the geometric mean of B1 = radical b19xb1 = radical B1 ^ 2xq ^ 18 = radical 1 / 4 x 2 ^ 9 = radical 2 ^ 7 = 8 * radical 2
(2010 Susong County three modules) arithmetic {an} series of the former n items and Sn, S9=-18, S13=-52, geometric ratio {bn}, b5=a5, b7=a7, the value of B15 is ()
A. 64B. -64C. 128D. -128
∵ S9 = 92 (a1 + A9) = 9A5 = - 18, S13 = 132 (a1 + A13) = 13a7 = - 52, ∵ A5 = - 2, a7 = - 4, and ∵ B5 = A5, B7 = A7, ∵ B5 = - 2, B7 = - 4, ∵ Q2 = 2, B15 = B7 · Q8 = - 4 × 16 = - 64
SN is the sum of the first n terms of the arithmetic sequence {an}, S9 = - 36, S13 = - 104, in the arithmetic sequence {BN}, B5 = A5, B7 = A7, then B6 is equal to ()
A. 42B. ±22C. ±42D. 32
In the arithmetic sequence, by using the properties of the middle term in the arithmetic, we get that S9 = 9 × a1 + a92 = 9 × A5 = - 36, A5 = - 4, S13 = 13 × (a1 + A13) × 12 = 13 × A7 = - 104 A7 = - 8. Because in the arithmetic sequence {BN}, B5 = A5, B7 = A7, so B6 = ± B5 · B7 = ± A5 · A7 = ± 42
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