HOW TO FIND THE NTH TERM OF A SEQUENCE

Write the nth term of the following sequences ;

Example 1 :

2, 2, 4, 4, 6, 6, ........

Solution :

By observing the given sequence first, second terms are same, third and fourth terms are same and so on.

To find nth term of the sequence, let us divide them into two parts.

Hence t_n = n + 1  (if n is odd)

               = n (if n is even)

Example 2 :

1/2 , 2/3 , 3/4 , 4/5 , 5/6, ........

Solution :

By observing the given sequence, the numerator is 1 lesser than the denominator.

Let t_n  =  n/(n+1) 

t₁  =  1/(1+1)  ==>  1/2

t₂  =  2/(2+1)  ==>  2/3

t₃  =  3/(3+1)  ==>  3/4

Hence the required nth term of the given sequence is n/(n+1).

Example 3 :

1/2, 3/4, 5/6, 7/8, 9/10, ........

Solution :

By observing the given sequence, the numerator is 1 lesser than the denominator and they are odd terms.

Let t_n  =  (2n-1)/2n

t₁  =  (2(1) - 1)/2(1) ==>  1/2

t₂  =  (2(2) - 1)/2(2) ==>  3/4

t₃  =  (2(3) - 1)/2(3) ==>  5/6

Hence the required nth term of the given sequence is (2n-1)/2n.

Example 4 :

6, 10, 4, 12, 2, 14, 0, 16, −2, ........

Solution :

By observing the given sequence first, second terms are same, third and fourth terms are same and so on.

To find nth term of the sequence, let us divide them into two parts.

Hence the nth term is 

t_n = 7 - n  (if n is odd) 

    = 8 + n  (if n is even)

Example 5 :

Kelly is saying her money to buy a car. She has $50000 and she plans to save $3750 per week from her job as a call center manager.

a) how much will Kelly have saved after 8 weeks ?

b)  If the car's down payment cost $270000, how long will it take her to save money at this rate ?

Solution :

Initial amount he has = $50000

a = 50000

Savings per week = 3750

Common difference (d) = 3750

a)

Saving in n weeks :

a_n = a + (n - 1) d

= 50000 + (n - 1) 3750

= 50000 + 3750n - 3750

a_n = 3750n + 46250

Savings, after 8 weeks

a₈ = 3750(8) + 46250

= 30000 + 46250

= 76250

b)  270000 = 3750n + 46250

270000 - 46250 = 3750n

223750 = 3750n

n = 223750/3750

n = 59.6

Approximately 60 weeks.

Example 6 :

There are 28 seats in the front row of a theater. Each successive row contains two more seat than the pervious row. If there are 24 row, how many seats are there in the last row ?

Solution :

Number of seats in the first row (a) = 28

Number of seats in the next row = 28 + 2 ==> 30 (two more seats)

d = 30

Number of rows = 24

Total number of seats :

S_n = (n/2) [2a + (n - 1) d]

S₂₄ = (24/2) [2(28) + (24 - 1)2]

= 12[56 + 23(2)]

= 12[56 + 46]

= 12(102)

= 1224

So, the total number seats is 1224.

Example 7 :

Mario began an exercise program to get back in shape. He plans to row 5 minutes on his rowing machine the first day and increase his rowing time by one minute and thirty seconds each day.

a) How long will he row on the 18th day ?

b) On what day will Mari first row and hour and more ?

Solution :

First day (a) = 5 minutes

increases by thirty seconds each day (d) = 1 minute 30 seconds

Converting into minutes, we get

= 1 + 30/60 minutes

= 3/2 minutes or 1.5 seconds

a_n = a + (n - 1) d

a_n = 5 + (n - 1) (1.5)

= 5 + 1.5n - 1.5

a_n = 3.5 + 1.5n

a)

Time taken he will row on 18th day :

a₁₈ = 3.5 + 1.5(18)

= 3.5 + 27

= 30.5

b) 3.5 + 1.5n > 60

1.5n > 60 - 3.5

1.5n > 56.5

n > 56.5/1.5

n > 37.6

Since the day number must be a whole number, Mario will first row for an hour or more on day 38. 

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