b1=1/2 n=7 Sn=127/128 q-? bn-?

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вопрос

To find the common ratio (q) and the 7th term (bn), we can use the formula for the sum of the first n terms of a geometric progression:

Sn = b1 * (1 - q^n) / (1 - q)

Given that Sn = 127/128, b1 = 1/2, and n = 7, we can substitute these values into the formula to find q:

127/128 = (1/2) * (1 - q^7) / (1 - q)

Multiplying both sides by (1 - q) and expanding, we get:

127/128 - (127/256)q = (1/2) - (1/2)q^7

Simplifying further, we get:

127/128 - 127/256q = 1/2 - 1/2q^7

Multiplying both sides by 256 to get rid of the fractions, we have:

256(127/128) - 127q = 128 - 128q^7

Solving for q, we get:

q = 1/2

Therefore, the common ratio (q) is 1/2.

To find the 7th term (bn), we can use the formula for the nth term of a geometric progression:

bn = b1 * q^(n-1)

Substituting the values of b1 = 1/2, q = 1/2, and n = 7 into the formula, we get:

b7 = (1/2) (1/2)^(7-1)
b7 = (1/2) (1/2)^6
b7 = (1/2) (1/2)^6
b7 = (1/2) (1/64)
b7 = 1/128

Therefore, the 7th term (b7) is 1/128.