# CAT Questions on AP GP HP

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CAT Questions on AP GP HP:

Download CAT Questions and Answers PDF on AP, GP & HP. Progressions and series is an important topic in CAT exam. Practice solved questions on Arithmetic, Harmonic & Geometric progression.

Question 1: Find the sum of all the 3-digit integers that leave a remainder of 5 when divided by 7.

a) 59489
b) 66879
c) 71079
d) 78659

Question 2: If 2x + y = 28, then find the maximum value of $x^4y^3$.

a) $3^{18}*2^3$
b) $3^{15}*2^5$
c) $2^{15}*3^5$
d) $2^{18}*3^3$

Question 3: Find the sum of all the odd integers between 150 to 246 that do not end in 3.

a) 9504
b) 7524
c) 8634
d) 8164

Question 4: If 0.43232323232… = $\frac{a}{b}$ such that both a and b are integers, then find the least value of a+b.

a) 635
b) 684
c) 709
d) 736

Question 5: Every term of the series starting from the third term is the sum of two preceding terms. If the first term is odd, the second term is even and the total number of terms in the series in 150, then find the ratio of number of even terms to the number of odd terms.

a) 1:3
b) 3:1
c) 2:1
d) 1:2

Question 6: If the first term of an infinite geometric progression is equal to twice the sum of terms that follow, then what is the ratio of the third term to the sixth term ?

a) 27 : 1
b) 1 : 27
c) 9 : 1
d) cannot be determined

Question 7: Find the sum to infinite terms of $\frac{1}{3^2-4}+\frac{1}{4^2-4}+\frac{1}{5^2-4}+\frac{1}{6^2-4}+…$

a) $\frac{33}{64}$
b) $\frac{35}{64}$
c) $\frac{25}{48}$
d) $\frac{23}{48}$

Question 8: Find the sum of the following series: 1 + 2 + 5 + 10 + 17 + 26….till 50 terms.
a) 43375
b) 40475
c) 42250
d) None of these

Question 9: A sequence is such that in any set of four consecutive terms, the sum of first and third term is equal to the sum of second and fourth term. If the third term is equal to 5, 26th term is equal to 9 and the sum of first 18 terms is equal to 58, find the first term.
a) 1
b) 2
c) 3
d) 4

Question 10: The internal angles of a convex polygon are in arithmetic progression with a common difference of 10 degrees. If the smallest angle is 100 degrees, what is the number of sides of the polygon?
a) 10
b) 7
c) 8
d) 9

The first three digit number that leaves a remainder of 5 when divided by 7 is 103
103, 110, 117, 124,….
This forms an Arithmetic Progression.
The last three digit number that leaves a remainder of 5 when divided by 7 is 999.
$T_n$ = first_term + (n-1)(common_difference)
999 = 103 + (n-1)7
=> 7(n-1) = 896
=> n-1 = 128
=> n = 129
Sum = $\frac{129}{2}*(103+999)$
= $\frac{129}{2}*(1102)$
= 129 * 551
= 71079

2x + y = 28 => ($\frac{x}{2}$) + ($\frac{x}{2}$) + ($\frac{x}{2}$) + ($\frac{x}{2}$) + ($\frac{y}{3}$) + ($\frac{y}{3}$) + ($\frac{y}{3}$) = 28

As AM>=GM
=> $\frac{(\frac{x}{2}) + (\frac{x}{2}) + (\frac{x}{2}) + (\frac{x}{2}) + (\frac{y}{3}) + (\frac{y}{3}) + (\frac{y}{3})}{7} \geq ((\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(\frac{y}{3})(\frac{y}{3})(\frac{y}{3})^{\frac{1}{7})}$
=> $\frac{28}{7} \geq ((\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(\frac{y}{3})(\frac{y}{3})(\frac{y}{3})^{\frac{1}{7})}$
=> $(\frac{x^4y^3}{2^4*3^3})^{\frac{1}{7}} \leq 4$
=> $\frac{x^4y^3}{2^4*3^3} \leq 2^{14}$
=> $x^4y^3 \leq 2^{18}*3^3$

Sum of all odd integers upto 245 = $123^2$
Sum of all odd integers upto 149 = $75^2$
=> Sum of all odd integers between 150 and 246 = $123^2-75^2$ = 198*48 = 9504\\
Numbers between 150 and 246 that end in 3 are 153, 163, 173, 183, 193, 203, 213, 223, 233 and 243
There are in Arithmetic Progression.
=> Sum = $\frac{10}{2}(153+243)$ = 1980
=> Required sum = 9504 – 1980 = 7524

0.43232323232… = 0.4 + 0.032 + 0.00032 + 0.0000032 + …
= $\frac{4}{10}$ + $\frac{32}{10^3}$ + $\frac{32}{10^5}$ + $\frac{32}{10^7}$ + …
= $\frac{4}{10}$ + $\frac{32}{10^3} * (1 + \frac{1}{10^2} + \frac{1}{10^4} + \frac{1}{10^6} + \frac{1}{10^8} + …)$
= $\frac{4}{10}$ + $\frac{32}{10^3} * (\frac{1}{1-\frac{1}{10^2}})$
= $\frac{4}{10}$ + $\frac{32}{10^3} * (\frac{100}{99})$
= $\frac{4}{10}$ + $\frac{32}{990}$
= $\frac{396+32}{990}$
= $\frac{428}{990}$
= $\frac{214}{495}$ = $\frac{a}{b}$
=> a + b = 214 + 495 = 709

The first term is odd and the second term is even.
O, E, O, O, E, O, O, E and so on.
It means the second term, the fifth term, the eigth term and so on till 149th term are even terms.
Number of even terms = x
149 = 2+(x-1)3
x = 50
Number of odd terms = 100
Ratio = 50:100 = 1:2

Let the first term of the infinite G.P be a and r the common ratio.
Given, a = 2(ar + ar$^2$ + ar$^3$ …..)
a = 2($\frac{ar}{1-r}$)
1 – r = 2r , thus r = 1/3
The ratio of the third term to the ratio of the sixth term = $\frac{ar^2}{ar^5} = \frac{27}{1}$
Thus, A is the right choice.

Let S = $\frac{1}{3^2-4}+\frac{1}{4^2-4}+\frac{1}{5^2-4}+\frac{1}{6^2-4}+…$
=> S = $\frac{1}{(3+2)(3-2)}+\frac{1}{(4+2)(4-2)}+\frac{1}{(5+2)(5-2)}+\frac{1}{(6+2)(6-2)}+…$
=> S = $\frac{1}{(1)(5)}+\frac{1}{(2)(6)}+\frac{1}{(3)(7)}+\frac{1}{(4)(8)}+…..$
=> S = $\frac{1}{4}(1-\frac{1}{5})+\frac{1}{4}(\frac{1}{2}-\frac{1}{6})+\frac{1}{4}(\frac{1}{3}-\frac{1}{7})+\frac{1}{4}(\frac{1}{4}-\frac{1}{8})+\frac{1}{4}(\frac{1}{5}-\frac{1}{9})+….$
=> S = $\frac{1}{4}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})$
=> s = $\frac{25}{48}$

S = 1 + 2 + 5 + 10 + 17 + 26 +….+ tn
S-S = 1 + (2-1) + (5-2) + (10-5) + (17-10) +….. +[tn-t(n-1)]-tn
0 = 1 + (1 + 3 + 5 + 7 +……+ (n-1) terms) – tn

=> tn = 1 + (n-1)/2 (2 + (n-2)2) = 1 + (n-1)( 1 + n-2) = 1 + $(n-1)^2$
=>tn = $(n^2 – 2n + 2)$
Sum = (n)(n+1)(2n+1)/6 – n(n+1) + 2n = 50*51*101/6 – 50*51 + 100 = 40475

Let the first term be a, second term be x and the third term be c.
Fourth term = a+c-x
Fifth term = a
Sixth term = x
In this way pattern will repeat.
26th term = 2nd term = x = 9
c = 5
2(a+c)*4+a+x = 58
9a+8c+x = 58
9a+40+9 = 58
a = 1

Therefore, $n^2 + 19n = 36n – 72$
So, $n^2 – 17n +72 = 0$
Hence, $n=8$ or $n=9$