CAT Questions on Arithmetic & Geometric Progression Set-2 PDF

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CAT Questions on Geometric Progression Set-2 PDF
CAT Questions on Geometric Progression Set-2 PDF

CAT Questions on Geometric Progression Set-2 PDF:

Download CAT Questions on Geometric Progression Set-2 PDF. Practice important CAT Questions on Geometric Progression Questions with detailed answers and explanations. These questions are based on previous CAT question papers.

Download CAT Questions on Geometric Progression Set-2 PDF

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Question 1: In a Geometric Progression, the sum of first 3x term of the series is S and the sum of first 2x terms of the series is 12S/133. If the sum of first x terms of the series is S/k, find the value of ‘k’. It is given that the common difference of the GP is positive.

a) 120
b) 133
c) 155
d) 160

Question 2: The sides of a right angled triangle are in Geometric Progression. What is the ratio of cos of the acute angles of the triangle ?

a) $\sqrt{5}/2$
b) $\sqrt{ (1 + \sqrt{5})/2}$
c) $(1 + \sqrt{5})/2$
d) $1$

Question 3: In an infinite geometric progression with common ratio less than 1 the sum of any two consecutive terms is 8 times the sum of all the terms that follow them. What is the ratio of any term and the sum of all the terms that follow it?

a) 2
b) -2
c) -4
d) Cannot be determined

Question 4: In an arithmetic progression having 30 terms, the sum of 13 terms is equal to the sum of 27 terms. Which of the following is necessarily true?

a) The sum of all the terms in the progression is 0.
b) The sum of all the terms of the progression is negative.
c) 14th term of the progression is 0.
d) None of these

Question 5: In an Arithmetic progression, the sum of first 10 terms is half the the sum of first 15 terms. Find the ratio of the sum of first 16 terms and first 21 terms of the same AP.

a) 7:11
b) 6:10
c) 12:17
d) 8:13

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Answers & Solutions:

1) Answer (B)

The sum of the first 3x terms is 12S/133. Let us assume S as 133 and x=1.
So, the sum of first 3 terms is 133, and the sum of first 2 terms is 12.
=> The third term is 133-12=121.
Let the terms be $a, ar, ar^2$.
=> $a+ar=12$ or $a(r+1)=12$ and $ar^2=121$
The only values of a and r that satisfy these equations is a=1 and r=11
So, the first term is 1 i.e. 133/133 or S/133. => k=133
Alternate Solution:
Assume ‘a’ as the first term of the GP and ‘r’ as the ratio.
=> $\frac{a(r^{3x}-1)}{r-1}=S$ —-(1)
$\frac{a(r^{2x}-1)}{r-1}=12S/133$ —(2)
$\frac{a(r^{x}-1)}{r-1}=S/k$ —-(3)
Dividing 1 by 2 we get-
$\frac{(r^{3x}-1)}{r^{2x}-1}=\frac{133}{12}$
=>$\frac{(r^{2x}+r^{x}+1)}{r^{x}+1}=\frac{133}{12}$
=>$\frac{r^{2x}}{r^{x}+1}=\frac{121}{12}$
Assuming $r^x=m$, we get $\frac{m^{2}}{m+1}=\frac{121}{12}$
Solving this we get 2 values of m. One is 11 and the other is negative. But since r>0, m can’t be negative.
=> $m=r^{x}=11$
Dividing 1 by 3 we get-
$\frac{(r^{3x}-1)}{r^{x}-1}=k$
=> $r^{2x}+r^{x}+1=k$
Putting $r^{x}=11$, we get k=133

2) Answer (B)

Let a, ar and ar$^2$ be the three sides of the right triangle.
Since, it is a right triangle, $(ar^2)^2 = a^2 + (ar)^2$
$r^4 = r^2 + 1$
$r^4 – r^2 – 1 = 0$
$r^2 = (1 \pm \sqrt{1 – (-4)})/2$
$r^2 = (1 \pm \sqrt{5})/2$
Since, $r^2$ > 0,
$r^2 = (1 + \sqrt{5})/2$
We have to find cosA/cosC or cosC/cosA
cosA = a/ar$^2$ = 1/r$^2$
cosC = ar/ar$^2$ = 1/r
cosC/cosA = r = $\sqrt{ (1 + \sqrt{5})/2}$.5) Answer (D)
Thus, option B is the right choice.

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3) Answer (D)

Let the terms of the GP be $a, ar, ar^2, ar^3, ….$
The sum of the first two terms = $a(1+r)$
The sum of all the following terms = $\frac{ar^2}{1-r}$
It is given that, $a(1+r)=8\times\frac{ar^2}{1-r}$
=> $1-r^2=8r^2$
=> $r=\frac{1}{3}or\frac{-1}{3}$.
The ratio of any term and the sum of all terms that follow is,
$\frac{a}{\frac{ar}{1-r}}=\frac{1-r}{r}$
The value of this is 2 or -4.
Hence the answer is Option D.

4) Answer (D)

The sum of all the terms will not be zero. The sum of 13 terms is equal to that of 27 terms. So this means that the sum of the terms from 14th to 27th is 0. But this does not mean that sum of all the terms is 0.
The sum of all the terms can either be positive or negative. It will depend on whether the common difference is positive or negative.
14th term will not be 0.
Thus none of the given statements is necessarily true.
Hence option d is the correct answer.

5) Answer (D)

We have been given that
$S_{10} = \frac{1}{2}S_{15}$
If the first term of the sequence is ‘a’ and the common difference is ‘d’ then
$\frac{15}{2}[2a + 14d] = 2*\frac{10}{2}[2a + 9d]$

=> 15a + 105d = 20a + 90 d
=> 5a = 15d
=> a = 3d
=> So $S_{16}$ = 16/2 [2a + 15d] = 16a + 120d = 16a + 40a = 56a
=>$S_{21}$ = 21/2[2a + 20d] = 21a + 210d => 21a + 70a = 91a
Hence the required ratio is 56a/91a = 56/91 = 8/13

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