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)^6b7 = (1/2) (1/2)^6b7 = (1/2) (1/64)b7 = 1/128
Therefore, the 7th term (b7) is 1/128.
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.