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If the first and the nth term of a G.P are a and b , respectively , and if P is the product of n terms , prove that P*2 = [ab]*n ?
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If the first and the nth term of a G.P are a and b , respectively , an...
Proof:
Given: First term of G.P = a and nth term of G.P = b
To prove: P*2 = [ab]*n
Step 1: Find the common ratio of the G.P
Let the common ratio of the G.P be r.
Then, we know that b = ar^(n-1)
Step 2: Find the product of the first and nth terms of the G.P
The product of the first and nth terms of the G.P is ab.
Step 3: Find the product of the G.P
The product of the G.P can be written as P = a * ar * ar^2 * ... * ar^(n-1) = a^n * r^(1+2+...+(n-1)) = a^n * r^(n*(n-1)/2)
Step 4: Simplify the expression for the product of the G.P
We know that r^(n-1) = b/a, so r^n = b/a * r.
Substituting this in the expression for P, we get:
P = a^n * r^(n*(n-1)/2) = a^n * (b/a)^((n-1)/2) * r^(n*(n-1)/2)
= b^((n-1)/2) * (a*r^(n-1))^((n-1)/2)
= b^((n-1)/2) * b^((n-1)/2)
= b^n
Step 5: Prove P*2 = [ab]*n
P*2 = 2b^n (from Step 4)
[ab]*n = ab * n * r^(n-1) = ab * n * (b/a) = b^n * n
Since P*2 = b^n * 2, we can see that P*2 = [ab]*n.
Hence, we have proved that P*2 = [ab]*n.
Given: First term of G.P = a and nth term of G.P = b
To prove: P*2 = [ab]*n
Step 1: Find the common ratio of the G.P
Let the common ratio of the G.P be r.
Then, we know that b = ar^(n-1)
Step 2: Find the product of the first and nth terms of the G.P
The product of the first and nth terms of the G.P is ab.
Step 3: Find the product of the G.P
The product of the G.P can be written as P = a * ar * ar^2 * ... * ar^(n-1) = a^n * r^(1+2+...+(n-1)) = a^n * r^(n*(n-1)/2)
Step 4: Simplify the expression for the product of the G.P
We know that r^(n-1) = b/a, so r^n = b/a * r.
Substituting this in the expression for P, we get:
P = a^n * r^(n*(n-1)/2) = a^n * (b/a)^((n-1)/2) * r^(n*(n-1)/2)
= b^((n-1)/2) * (a*r^(n-1))^((n-1)/2)
= b^((n-1)/2) * b^((n-1)/2)
= b^n
Step 5: Prove P*2 = [ab]*n
P*2 = 2b^n (from Step 4)
[ab]*n = ab * n * r^(n-1) = ab * n * (b/a) = b^n * n
Since P*2 = b^n * 2, we can see that P*2 = [ab]*n.
Hence, we have proved that P*2 = [ab]*n.
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If the first and the nth term of a G.P are a and b , respectively , and if P is the product of n terms , prove that P*2 = [ab]*n ?
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If the first and the nth term of a G.P are a and b , respectively , and if P is the product of n terms , prove that P*2 = [ab]*n ? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about If the first and the nth term of a G.P are a and b , respectively , and if P is the product of n terms , prove that P*2 = [ab]*n ? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the first and the nth term of a G.P are a and b , respectively , and if P is the product of n terms , prove that P*2 = [ab]*n ?.
If the first and the nth term of a G.P are a and b , respectively , and if P is the product of n terms , prove that P*2 = [ab]*n ? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about If the first and the nth term of a G.P are a and b , respectively , and if P is the product of n terms , prove that P*2 = [ab]*n ? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the first and the nth term of a G.P are a and b , respectively , and if P is the product of n terms , prove that P*2 = [ab]*n ?.
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