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QUESTION: 1

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

Solution:

Number of terms = { (1st term - last term) / common difference} + 1

= {(140-20) / 5} + 1

⇒ (120/5) + 1

⇒ 24 + 1 = 25.

QUESTION: 2

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

Solution:

__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.

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.

QUESTION: 3

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

Solution:

15^{th} term = a + 14d

⇒ 20 + 14*(-5)

⇒ 20 - 70 = -50.

QUESTION: 4

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

Solution:

__1 ^{st} Method:__

1

3

Then, d = 5

16

Sum = {n*(a+L)/2} = {No. of terms*(first term + last term)/2}.

Thus, sum = {16*(5+80)/2} = 680.

Sum = Number of terms * Average of that AP.

Sum = 16* {(5+80)/2} = 16*45 = 680.

QUESTION: 5

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

Solution:

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

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?

Solution:

1^{st} term = 1

Common ratio = 2

Sum (S_{n}) = a*(r^{n}-1)/(r-1)

⇒ 1*(2^{20}-1)/(2-1)

⇒ 2^{20}-1.

QUESTION: 7

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

Solution:

5^{th} term of GP = ar^{5-1}

⇒ 16*r^{4} = 81

⇒ r = (81/16)^{1/4} = 3/2

4^{th} term of GP = ar^{4-1}

⇒ 16*(3/2)^{3} = 54.

QUESTION: 8

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

Solution:

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.

QUESTION: 9

After striking the floor, a rubber ball rebounds to 4/5^{th} 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.

Solution:

► 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.

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 :

Solution:

Total number of bacteria after 10 seconds,

⇒ 3^{10} - 3^{5}

⇒ 3^{5} *(3^{5} -1)

⇒ 243 *(3^{5} -1)

Since, just after 10 seconds all the bacterias (i.e. 3^{5} ) are dead after living 5 seconds each.

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