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Test: Progression (AP And GP)- 1


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

How many terms are there in 20, 25, 30......... 140

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

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

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

Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.

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

1st 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.

2nd 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.

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

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

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

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

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

The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is

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

1st Method:
1st term = 5
3rd term = 15
Then, d = 5
16th 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.

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

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

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

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

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

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

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 6

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 7

If the fifth term of a GP is 81 and first term is 16, what will be the 4th term of the GP?

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

5th term of GP = ar5-1 
⇒ 16*r4 = 81
⇒ r = (81/16)1/4 = 3/2
4th term of GP = ar4-1 
⇒ 16*(3/2)3 = 54.

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

The 7th and 21st terms of an AP are 6 and -22 respectively. Find the 26th term.

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

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

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

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 9

► 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 10

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 10

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.

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