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

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

Let term = l = ar^{n - 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

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

► 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.

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.

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

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

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

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.

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