Progressions (January 15) - CAT MCQ

Number of terms = { (1st term - last term) / common difference} + 1
= {(140-20) / 5} + 1
⇒ (120/5) + 1
⇒ 24 + 1 = 25.

1^st Method:
8th term = a+7d = 39 ......(i)
12th term = a+11d = 59 ......(ii)
Subtracting (ii) from (i) we get,
a + 7d - a - 11d = 39 - 59 
⇒ 4d = 20 or d = 5
Hence, a + 7*5 = 39
Thus, a = 39 - 35 = 4.

2^nd Method (Thought Process):
8th term = 39
And, 12th term = 59
Here, we see that 20 is added to 8th term 39 to get 12th term 59 i.e. 4 times the common difference is added to 39.
So, CD = 20/4 = 5.
Hence, 7 times CD is added to 1st term to get 39. That means 4 is the 1st term of the AP.

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

1^st Method:
1^st term = 5
3^rd term = 15
Then, d = 5
16^th term = a + 15d = 5 + 15*5 = 80
Sum = {n*(a+L)/2} = {No. of terms*(first term + last term)/2}.
Thus, sum = {16*(5+80)/2} = 680.

2^nd Method (Thought Process): 
Sum = Number of terms * Average of that AP.
Sum = 16* {(5+80)/2} = 16*45 = 680.

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.

5^th term of GP = ar^5-1 
⇒ 16*r⁴ = 81
⇒ r = (81/16)^1/4 = 3/2
4^th term of GP = ar^4-1 
⇒ 16*(3/2)³ = 54.

7^th term = 6
21^st term = -22
That means, 14 times common difference or -28 is added to 6 to get -22
Thus, d = -2
7^st term = 6 = a+6d
⇒ a + (6*-2) = 6
⇒ a = 18
26^st term = a + 25d
⇒ 18 - 25*2 = -32.

► 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.