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Test: Progression (AP And GP)- 1 - CAT MCQ


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10 Questions MCQ Test Quantitative Aptitude (Quant) - Test: Progression (AP And GP)- 1

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Test: Progression (AP And GP)- 1 - Question 1

Find the 15th term of the sequence 20, 15, 10....

Detailed Solution for Test: Progression (AP And GP)- 1 - Question 1

15th term = a + 14d
⇒ 20 + 14*(-5)
⇒ 20 - 70 = -50.

Test: Progression (AP And GP)- 1 - Question 2

How many terms are there in the GP 5, 20, 80, 320........... 20480?

Detailed Solution for Test: Progression (AP And GP)- 1 - Question 2

Let term = l = arn - 1 a = 5 and l = 20480
r = 20/5 = 4
∴ 20480 = 5 x (4)n-1
(4)n-1 = 20480/5 = 4096 = (4)6
n - 1 = 6
∴ n = 7

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Test: Progression (AP And GP)- 1 - Question 3

A boy agrees to work at the rate of one rupee on the first day, two rupees on the second day, and four rupees on third day and so on. How much will the boy get if he started working on the 1st of February and finishes on the 20th of February?

Detailed Solution for Test: Progression (AP And GP)- 1 - Question 3

1st term = 1
Common ratio = 2
Sum (Sn) = a*(rn-1)/(r-1)
⇒ 1*(220-1)/(2-1)
⇒ 220-1.

Test: Progression (AP And GP)- 1 - Question 4

After striking the floor, a rubber ball rebounds to 4/5th of the height from which it has fallen. Find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 metres.

Detailed Solution for Test: Progression (AP And GP)- 1 - Question 4

► So, starting from a height of 120m, the object will rebound to 4/5th of its original height after striking the floor each and every time.
► This means after the first drop the ball will rebound and will fall with that rebound metres, and so on.

► We get a series from these observations

Total distance travelled = 120 + 96 + 96 + 76.8 + 76.8 + 61.4 + 61.4 + ….

⇒ 120 + 2*(96 + 76.8 + 61.4 …)

⇒ 120 + 2*(96 + 96*(4/5) + 96(4/5)2 + …)

► Now, inside those brackets, it is a Geometric Progression or a GP with the first term a = 96 and the common ratio r = 4/5 = 0.8

► As it is an infinite GP, the sum of all it's terms is equal to a / (1-r)

So, the sum of distances covered = 120 + 2*96 / (1 - 0.8)
⇒ 120 + 960 = 1080m.

Test: Progression (AP And GP)- 1 - Question 5

A bacteria gives birth to two new bacteria in each second and the life span of each bacteria is 5 seconds. The process of the reproduction is continuous until the death of the bacteria. initially there is one newly born bacteria at time t = 0, the find the total number of live bacteria just after 10 seconds :

Detailed Solution for Test: Progression (AP And GP)- 1 - Question 5

Total number of bacteria after 10 seconds,
⇒ 310 - 35
⇒ 35 *(35 -1)
⇒ 243 *(35 -1)
Since, just after 10 seconds all the bacterias (i.e. 35 ) are dead after living 5 seconds each.

Test: Progression (AP And GP)- 1 - Question 6

Three positive integers x, y and z are in arithmetic progression. If y −x > 2 xyz = 5(x+y+z), then z-x equals ?

Detailed Solution for Test: Progression (AP And GP)- 1 - Question 6

Given x, y, z are three terms in an arithmetic progression.
Considering x = a, y = a+d, z = a+2*d.
Using the given equation x*y*z = 5*(x+y+z)  
a*(a+d)*(a+2*d) = 5*(a+a+d+a+2*d)
=a*(a+d)*(a+2*d) 
= 5*(3*a+3*d)
= 15*(a+d).
= a*(a+2*d)
= 15.
Since all x, y, z are positive integers and y-x > 2. a, a+d, a+2*d are integers.
The common difference is positive and greater than 2.
Among the different possibilities are : (a=1, a+2d = 5), (a, =3, a+2d = 5), (a = 5, a+2d = 3), (a=15, a+2d = 1)
Hence the only possible case satisfying the condition is : a = 1, a+2*d = 15. x = 1, z = 15. z-x = 14

Test: Progression (AP And GP)- 1 - Question 7

If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms?

Detailed Solution for Test: Progression (AP And GP)- 1 - Question 7

Sum of the first 11 terms = 11/2 ( 2a+10d)

Sum of the first 19 terms = 19/2 (2a+18d)

=> 22a+110d = 38a+342d => 16a =- 232d

=> 2a = -232/8 d = -29d

Sum of the first 30 terms = 15(2a+29d) = 0

Test: Progression (AP And GP)- 1 - Question 8

If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be

Detailed Solution for Test: Progression (AP And GP)- 1 - Question 8

The population of town at the beginning of 1st year = p
The population of town at the beginning of 2nd year = 3+2p
The population of town at the beginning of 3rd year = 2(3+2p)+3 = 2*2p+2*3+3 =4p+3(1+2)
The population of town at the beginning of 4th year = 2(2*2p+2*3+3)+3 = 8p+3(1+2+4)
The population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be =215(1003) - 3

Test: Progression (AP And GP)- 1 - Question 9

The number of common terms in the two sequences 17, 21, 25,…, 417 and 16, 21, 26,…, 466 is

Detailed Solution for Test: Progression (AP And GP)- 1 - Question 9

The terms of the first sequence are of the form 4p + 13
The terms of the second sequence are of the form 5q + 11
If a term is common to both the sequences, it is of the form 4p+13 and 5q+11
or 4p = 5q -2. LHS = 4p is always even, so, q is also even.
or 2p = 5r - 1 where q = 2r.
Notice that LHS is again even, hence r should be odd. Let r = 2m+1 for some m.
Hence, p = 5m + 2.
So, the number = 4p+13 = 20m + 21.
Hence, all numbers of the form 20m + 21 will be the common terms. i.e 21,41,61, ... ,401 = 20.

Test: Progression (AP And GP)- 1 - Question 10

If (2n +1)+(2n+3)+(2n+5)+...+(2n+47) = 5280 , then whatis the value of n ≥2 1 +2+3+ .. + n?


Detailed Solution for Test: Progression (AP And GP)- 1 - Question 10

Let us rst nd the number of terms 
47=1+(n-1)2 n=24 24*2n+1+3+5+....47=5280
48n+576=5280 48n=4704 n=98
Sum of first 98 terms = 98*99/2 =4851

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