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Given that (m +1l)th, (n + 1)th and (r +1l)th term of an AP are in GP and m , n , r are in HP, then find the ratio of the first term of the AP to its common difference in terms of n.
- a)2 : n
- b)n : 2
- c)2n : 3
- d)3 : 4n
Correct answer is option 'B'. Can you explain this answer?
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Given that (m +1l)th, (n + 1)th and (r +1l)th term of an AP are in GP ...
Since the (m + 1)th, (n + 1)th and (r + 1)th term of an A.P are in G.P. so,
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Given that (m +1l)th, (n + 1)th and (r +1l)th term of an AP are in GP ...
Given, (m+1)th, (n-1)th and (r+1)th term of an AP are in GP and m, n, r are in HP.
Let the first term of the AP be a and the common difference be d.
Then, we have:
(m+1)th term: a + md + d
(n-1)th term: a + (n-2)d
(r+1)th term: a + rd + d
Since the (m+1)th, (n-1)th, and (r+1)th terms are in GP, we have:
(a + md + d)(a + rd + d) = (a + (n-2)d)^2
Expanding and simplifying, we get:
a^2 + (m+n+r)ad + (mnr + 2m + 2n + 2r - 4) d^2 = a^2 + (n-2)^2 d^2
Cancelling out the a^2 terms and using the fact that m, n, and r are in HP, we get:
(m+n+r)ad = (n-2)^2 d^2
Simplifying, we get:
a/d = (n-2)/(m+n+r)
Since we need to express a/d in terms of n, we substitute m and r in terms of n using the fact that they are in HP:
2n = m + r
r = (2n^2)/(n-1)
m = (2n)/(n-1)
Substituting these values, we get:
a/d = (n-2)/(m+n+r)
= (n-2)/(2n/(n-1) + n + 2n^2/(n-1))
= n/(2n+1)
Therefore, the ratio of the first term of the AP to its common difference in terms of n is n:2. Thus, option B is the correct answer.
Let the first term of the AP be a and the common difference be d.
Then, we have:
(m+1)th term: a + md + d
(n-1)th term: a + (n-2)d
(r+1)th term: a + rd + d
Since the (m+1)th, (n-1)th, and (r+1)th terms are in GP, we have:
(a + md + d)(a + rd + d) = (a + (n-2)d)^2
Expanding and simplifying, we get:
a^2 + (m+n+r)ad + (mnr + 2m + 2n + 2r - 4) d^2 = a^2 + (n-2)^2 d^2
Cancelling out the a^2 terms and using the fact that m, n, and r are in HP, we get:
(m+n+r)ad = (n-2)^2 d^2
Simplifying, we get:
a/d = (n-2)/(m+n+r)
Since we need to express a/d in terms of n, we substitute m and r in terms of n using the fact that they are in HP:
2n = m + r
r = (2n^2)/(n-1)
m = (2n)/(n-1)
Substituting these values, we get:
a/d = (n-2)/(m+n+r)
= (n-2)/(2n/(n-1) + n + 2n^2/(n-1))
= n/(2n+1)
Therefore, the ratio of the first term of the AP to its common difference in terms of n is n:2. Thus, option B is the correct answer.
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Given that (m +1l)th, (n + 1)th and (r +1l)th term of an AP are in GP and m , n , r are in HP, then find the ratio of the first term of the AP to its common difference in terms of n.a)2 : nb)n : 2c)2n : 3d)3 : 4nCorrect answer is option 'B'. Can you explain this answer?
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