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In a GP the sum of the first six terms is 28 times the sum of the first three terms. If the fourth term is 108, then find the second term in the series?
- a)-12
- b)-4
- c)-12
- d)-4
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
In a GP the sum of the first six terms is 28 times the sum of the fir...
Let's denote the first term of the geometric progression (GP) as 'a' and the common ratio as 'r'.
The sum of the first n terms of a GP can be calculated using the formula:
Sn = a * (1 - r^n) / (1 - r)
Given that the sum of the first six terms is 28 times the sum of the first three terms, we can write the following equation:
a * (1 - r^6) / (1 - r) = 28 * [a * (1 - r^3) / (1 - r)]
Simplifying this equation, we get:
(1 - r^6) / (1 - r) = 28 * (1 - r^3) / (1 - r)
Cross-multiplying and simplifying further, we have:
1 - r^6 = 28 - 28r^3
Adding 28r^3 to both sides and rearranging the terms, we get:
r^6 - 28r^3 + 27 = 0
Now, we know that the fourth term of the GP is 108. Therefore, we can write:
ar^3 = 108
Substituting this value of ar^3 into the equation above, we get:
(108)^2 - 28(108) + 27 = 0
Solving this quadratic equation, we find two possible values for r^3: 1 and 27/4.
If r^3 = 1, then r = 1, which means it is a common ratio of 1. However, this would make all the terms in the GP the same, which is not possible since the fourth term is given as 108.
Therefore, we consider the case where r^3 = 27/4. Solving this equation, we find that r = 3/2.
Now, we can use the formula for the nth term of a GP:
an = ar^(n-1)
Substituting the values of a (unknown) and r (3/2), and considering the fourth term (108), we can solve for the second term:
a * (3/2)^(4-1) = 108
a * (3/2)^3 = 108
a * (27/8) = 108
a = 108 * (8/27)
a = 32
Therefore, the second term in the series is given by:
a * r = 32 * (3/2) = 48
Hence, the correct answer is option C) -12.
The sum of the first n terms of a GP can be calculated using the formula:
Sn = a * (1 - r^n) / (1 - r)
Given that the sum of the first six terms is 28 times the sum of the first three terms, we can write the following equation:
a * (1 - r^6) / (1 - r) = 28 * [a * (1 - r^3) / (1 - r)]
Simplifying this equation, we get:
(1 - r^6) / (1 - r) = 28 * (1 - r^3) / (1 - r)
Cross-multiplying and simplifying further, we have:
1 - r^6 = 28 - 28r^3
Adding 28r^3 to both sides and rearranging the terms, we get:
r^6 - 28r^3 + 27 = 0
Now, we know that the fourth term of the GP is 108. Therefore, we can write:
ar^3 = 108
Substituting this value of ar^3 into the equation above, we get:
(108)^2 - 28(108) + 27 = 0
Solving this quadratic equation, we find two possible values for r^3: 1 and 27/4.
If r^3 = 1, then r = 1, which means it is a common ratio of 1. However, this would make all the terms in the GP the same, which is not possible since the fourth term is given as 108.
Therefore, we consider the case where r^3 = 27/4. Solving this equation, we find that r = 3/2.
Now, we can use the formula for the nth term of a GP:
an = ar^(n-1)
Substituting the values of a (unknown) and r (3/2), and considering the fourth term (108), we can solve for the second term:
a * (3/2)^(4-1) = 108
a * (3/2)^3 = 108
a * (27/8) = 108
a = 108 * (8/27)
a = 32
Therefore, the second term in the series is given by:
a * r = 32 * (3/2) = 48
Hence, the correct answer is option C) -12.
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Community Answer
In a GP the sum of the first six terms is 28 times the sum of the fir...
Let a be the first term and r be the common ratio.
On rearranging we get, 27 = r3(28 - r3)...(2)
also given, ar3 = 108
substituting (3) in (2)
On solving we get a = 4 and 108.
a ≠ = 108 as then we will get r = ± 1 which does not satisfy the condition given in (1) thus a = 4 and r = 3 (r ≠ = -3 as it does not satisfy (1))
∴ the second term = ar = 12
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In a GP the sum of the first six terms is 28 times the sum of the first three terms. If the fourth term is 108, then find the second term in the series?a)-12b)-4c)-12d)-4Correct answer is option 'C'. Can you explain this answer?
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