Let \(a,b,c,d \) be an increasing sequence of real numbers,which are in geometric progression .If \(a+d=112\) and \(b+c=48\) ,then the value of \(\dfrac{a+c+8}{b}\) is
\(1\)
\(5\)
\(4\)
\(3\)
\(2\)
The Correct Option is C
Solution and Explanation
Given that:
\(a + d = 112 \)
\(b + c = 48\)
\( a + ar^3 = 112 \)
\( ⇒a (1 + r^3) = 112\)-------(1)
\( ar + ar2 = 48 \)
\(⇒a ( r + r^2 ) = 48 \)--------(2)
Now comparing \(a\) from above cases (1) and (2)we get:
\(3r^3-7r^2-7r+3=0\)
on solving we get :
\(r= -1,3,0.3334\)
Therefore from parent equation we get \(a=\dfrac{48}{12}=4\)
Then, sequence becomes \( 4, 12, 36, 108\)\(\)
Therefore , \(\dfrac{a + c +8}{ b} = \dfrac{4+36+8}{12} = 4\) (_Ans)
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Concepts Used:
Geometric Progression
What is Geometric Sequence?
A geometric progression is the sequence, in which each term is varied by another by a common ratio. The next term of the sequence is produced when we multiply a constant to the previous term. It is represented by: a, ar1, ar2, ar3, ar4, and so on.
Properties of Geometric Progression (GP)
Important properties of GP are as follows:
- Three non-zero terms a, b, c are in GP if b2 = ac
- In a GP,
Three consecutive terms are as a/r, a, ar
Four consecutive terms are as a/r3, a/r, ar, ar3 - In a finite GP, the product of the terms equidistant from the beginning and the end term is the same that means, t1.tn = t2.tn-1 = t3.tn-2 = …..
- If each term of a GP is multiplied or divided by a non-zero constant, then the resulting sequence is also a GP with a common ratio
- The product and quotient of two GP’s is again a GP
- If each term of a GP is raised to power by the same non-zero quantity, the resultant sequence is also a GP.
If a1, a2, a3,… is a GP of positive terms then log a1, log a2, log a3,… is an AP (arithmetic progression) and vice versa