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Chapter 8: Problem 596

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) \(-8,-4,-2,-8\)

Short Answer

Expert verified
The short answer is (D) \(-8, -4, -2, -8\).

Step by step solution

01

Write the relationship between the numbers in G.P. and A.P.

For the G.P., we have general form a, ar, ar^2. Since a, b, and c are in G.P: \(a = a\) \(b = ar\) \(c = ar^2\) For the A.P., since c, d, and d+6 are in A.P: \(c = d\) \(d = c + 6\) (the common difference is 6) \(a = d\) (given in the problem)
02

Express all the terms in terms of a

Since a = d, rewrite the A.P. as: \(c = a\) \(d = a - 6\) \(a = a\) Now, the G.P. becomes: \(a = a\) \(b = a\div r\) \(c = a(1\div r)\)
03

Use the A.P. equations in the G.P. equations and solve for a

Substitute the value of c from A.P. into the G.P. equation for c: \(a = a(1\div r)\) Divide both sides by a (a≠0), we get: \(r = 1\) Substitute r back into the equation for b in G.P.: \(b = a\) Now we have: (1) \(a = a\) (2) \(b = a\) (3) \(c = a\)
04

Verify the answer by comparing with the given options

Using the information that a = b = c = d, let's compare our findings with the given options: (A) 8, 4, 2, 8: Incorrect, because b is not equal to a (B) -8, 4, -2, -8: Incorrect, because b is not equal to a (C) 8, -4, 2, 8: Incorrect, because b is not equal to a (D) -8, -4, -2, -8: Correct, because b = a, c = a, and d = a The correct answer is (D) \(-8, -4, -2, -8\).

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Most popular questions from this chapter

If the sum of first 101 terms of an A. P. is 0 and If 1 be the first term of the A. P. then the sum of next 100 terms is \((\mathrm{A})-101\) (B) 201 (C) \(-201\) (D) \(-200\)

In a G. P., the last term is 1024 and the common ratio is \(2 .\) Its 20 th term from the end is (A) \([1 /(512)]\) (B) \([1 /(1024)]\) (C) \([1 /(256)]\) (D) 512

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) \(\leq 9\) \((\mathrm{C})>9\) \((\mathrm{D}) \geq 9\)

\(\left(3 / 1^{2}\right)+\left[5 /\left(1^{2}+2^{2}\right)\right]+\left[7 /\left(1^{2}+2^{2}+3^{2}\right)\right] \ldots\) upto n terms \(-\) (A) \(\left[\left(6 \mathrm{n}^{2}\right) /(\mathrm{n}+1)\right]\) (B) \([(6 n) /(\mathrm{n}+1)]\) (C) \([\\{6(2 \mathrm{n}-1)\\} /(\mathrm{n}+1)]\) (D) \(\left[\left\\{3\left(n^{2}+1\right)\right\\} /(n+1)\right]\)

Find \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) between 2 and 18 such that \(\mathrm{a}+\mathrm{b}+\mathrm{c}=25\), \(2, a, b\) are in A. P. and \(b, c, 18\) are in G. P. (A) \(5,8,12\) (B) \(4,8,13\) (C) \(3,9,13\) (D) \(5,9,11\)
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