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Chapter 8: Problem 635

First term of a G. P. of \(2 \mathrm{n}\) terms is \(\mathrm{a}\), and the last term is 1 . The product of all the terms of the G. P. is (A) \((\mathrm{a} \ell)(\mathrm{n} / 2)\) (B) \((\mathrm{a} \ell)^{(\mathrm{n}-1)}\) (C) \((\mathrm{a} \ell)^{\mathrm{n}}\) (D) \((\mathrm{a} \ell)^{2 \mathrm{n}}\)

Short Answer

Expert verified
The short answer to the problem is: The product of all the terms of the G.P. is \((a^n)^{(2n-1)}\), which corresponds to option (B).

Step by step solution

01

Find the common ratio

In a G.P, the terms are formed by multiplying the previous term by a constant value called the common ratio (r). Since we know the first term is 'a' and the last term is 1, we can use the formula for the nth term of a G.P.: Last term = First term * (Common ratio)^(Number of terms - 1) This gives us: 1 = a * r^(2n - 1) Now, we need to find the common ratio, r, from the equation above.
02

Rearrange the equation to solve for r

To solve for r, we can rearrange the equation: r^(2n - 1) = 1/a Taking the (2n - 1)th root on both sides, we get: r = (1/a)^{1/(2n - 1)}
03

Calculate the product of all terms in the G.P.

The formula for the product of all terms in a G.P. is given by: Product of terms = (First term)^(Number of terms) * (Common ratio)^(Sum of all exponents in the series) Since there are 2n terms in the series, the sum of all exponents is Sum of exponents = 1 + 2 + ... + 2n - 1 = n(2n - 1) Thus, the product of terms in the G.P. is: Product of terms = a^(2n) * (1/a)^{n(2n - 1)} Simplifying further, we get: Product of terms = (a^n)^{2n - 1}
04

Match the result with the given options

Our result is (a^n)^{(2n-1)}, which corresponds to option (B): (B) \((\mathrm{a} \ell)^{(\mathrm{n}-1)}\) Therefore, the correct option is (B).

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Most popular questions from this chapter

If \(1, \mathrm{~A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3}, \ldots \mathrm{A}_{\mathrm{n}}, 31\) are in A. P. and \(\mathrm{A}_{7}: \mathrm{A}_{\mathrm{n}-1}=5: 9\) then \(\mathrm{n}=\) (A) 28 (B) 14 (C) 15 (D) 13

The sum of the series \((3 / 4)+(5 / 36)+[7 /(144)], \ldots\) up to 11 terms is (A) \([(120) \overline{/(121)]}\) (B) \([(143) /(144)]\) (C) 1 (A) \([(144) /(143)]\)

Find out four numbers such that, first three numbers are in G. P., last three numbers are in A. P. having common difference 6, first and last numbers are same. (A) \(8,4,2,8\) (B) \(-8,4,-2,-8\) (C) \(8,-4,2,8\) (D) \(-8,-4,-2,-8\)

If an A. P., \(\mathrm{T}_{35}=-50\) and \(\mathrm{d}=-3\) then \(\mathrm{S}_{35}=\) (A) 35 (B) 38 (C) 32 (D) 29

For all \(x, y \in R^{+}\) the value of \(\left[\left\\{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)\right\\} /(x y)\right]\) \(=\) (A) \(<9\) (B) \(\leq 9\) \((\mathrm{C})>9\) \((\mathrm{D}) \geq 9\)
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