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A GP consists of 1000 terms. The sum of the terms occupying the odd places is Pj and the sum of the terms occupying the even places is P2. Find the common ratio of this GP.
- a)P2/P1
- b)P1/P2
- c)(P2 - P1)/P1
- d)(P2+ P1)/P2
Correct answer is option 'A'. Can you explain this answer?
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A GP consists of 1000 terms. The sum of the terms occupying the odd pl...
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A GP consists of 1000 terms. The sum of the terms occupying the odd pl...
Given: A GP with 1000 terms, sum of terms in odd places = P1, and sum of terms in even places = P2
To find: Common ratio of the GP
Solution:
Let the first term of the GP be 'a' and the common ratio be 'r'. Then, the second term will be ar, the third term will be ar^2, and so on.
The sum of the terms in the odd places is given by:
P1 = a + ar^2 + ar^4 + ... + ar^998
Multiplying by r on both sides, we get:
P1r = ar + ar^3 + ar^5 + ... + ar^999
Subtracting the second equation from the first, we get:
P1(1 - r) = a - ar^999
Since the GP has 1000 terms, we know that a + ar + ar^2 + ... + ar^999 = a(1 - r^1000)/(1 - r). Therefore:
a - ar^999 = a(1 - r^1000)/(1 - r) - ar^999
Simplifying:
a(1 - r^1000)/(1 - r) = P1(1 - r)
a = P1(1 - r)/(1 - r^1000)
Similarly, the sum of the terms in the even places is given by:
P2 = ar + ar^3 + ar^5 + ... + ar^999
Multiplying by r on both sides, we get:
P2r = ar^2 + ar^4 + ar^6 + ... + ar^1000
Subtracting the second equation from the first, we get:
P2(1 - r) = ar - ar^1000
Substituting the value of 'a' from above:
P2(1 - r) = P1r(1 - r^999)/(1 - r^1000) - P1r^999
Simplifying:
P2 = P1r(1 - r^999)/(1 - r^1000)
Dividing P2 by P1, we get:
P2/P1 = r(1 - r^999)/(1 - r^1000)
To find the common ratio, we need to solve the above equation for 'r'. This can be done numerically or by making some assumptions and approximations. One such assumption is that the common ratio is small (i.e., r < 1).="" in="" this="" case,="" we="" can="" simplify="" the="" equation="" as="" />
P2/P1 ≈ r
Therefore, the common ratio of the GP is approximately equal to P2/P1. This is the correct answer, as given in option A.
To find: Common ratio of the GP
Solution:
Let the first term of the GP be 'a' and the common ratio be 'r'. Then, the second term will be ar, the third term will be ar^2, and so on.
The sum of the terms in the odd places is given by:
P1 = a + ar^2 + ar^4 + ... + ar^998
Multiplying by r on both sides, we get:
P1r = ar + ar^3 + ar^5 + ... + ar^999
Subtracting the second equation from the first, we get:
P1(1 - r) = a - ar^999
Since the GP has 1000 terms, we know that a + ar + ar^2 + ... + ar^999 = a(1 - r^1000)/(1 - r). Therefore:
a - ar^999 = a(1 - r^1000)/(1 - r) - ar^999
Simplifying:
a(1 - r^1000)/(1 - r) = P1(1 - r)
a = P1(1 - r)/(1 - r^1000)
Similarly, the sum of the terms in the even places is given by:
P2 = ar + ar^3 + ar^5 + ... + ar^999
Multiplying by r on both sides, we get:
P2r = ar^2 + ar^4 + ar^6 + ... + ar^1000
Subtracting the second equation from the first, we get:
P2(1 - r) = ar - ar^1000
Substituting the value of 'a' from above:
P2(1 - r) = P1r(1 - r^999)/(1 - r^1000) - P1r^999
Simplifying:
P2 = P1r(1 - r^999)/(1 - r^1000)
Dividing P2 by P1, we get:
P2/P1 = r(1 - r^999)/(1 - r^1000)
To find the common ratio, we need to solve the above equation for 'r'. This can be done numerically or by making some assumptions and approximations. One such assumption is that the common ratio is small (i.e., r < 1).="" in="" this="" case,="" we="" can="" simplify="" the="" equation="" as="" />
P2/P1 ≈ r
Therefore, the common ratio of the GP is approximately equal to P2/P1. This is the correct answer, as given in option A.
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A GP consists of 1000 terms. The sum of the terms occupying the odd places is Pj and the sum of the terms occupying the even places is P2. Find the common ratio of this GP.a)P2/P1b)P1/P2c)(P2 - P1)/P1d)(P2+ P1)/P2Correct answer is option 'A'. Can you explain this answer?
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