# In which field is arithmetic progression applied

DEFINITION

An arithmetic progression (AP) is a sequence or series of numbers called terms in which any term after the first can be obtain from its immediate predecessor by adding a fixed number called the “common difference” or “constant difference” denoted as “d”. The common difference can be positive or negative. When the common difference is positive, the AP is increasing and when it is negative, the AP is decreasing. For example 3, 7, 11, 15 etc is an AP since the common difference between one term and the next is 4.

Given, a as the first term

a + d the 2nd term

a + 2d the 3rd term,

From the above, the nth term denoted by Tn is given as Tn = a + (n-1) d

An arithmetic series is the sum of the terms of an AP.

Sum of an Arithmetic Progression

Where:

Sn is the sum of the nth term of an AP.

N.B: When the 1st term and the common difference of an AP are known, the AP is completely defined.

Application of AP

Example:

Mr. Teghen wants to embark in a saving scheme. He decides that the minimum amount he should save in any one year should be 100,000FCFA. At the end of the first year, he serves 100,000frs and save an additional 5000frs each succeeding year. How much shall he have save by the end of the 30th year.

Solution

a = 100000frs

d = 5000frs

n = 30

Example II( G.C.E Question: June 1999,Paper 2,Q20)

Miss “Hear Go Talk” begins a saving scheme and these she does in an AP. She finds out that after making 20 savings she had 1050000FCFA in her account an after 40 savings she had accumulated 4.100.000FCFA.

a)           Calculate her initial saving

b)          Calculate her incremental saving

c)           Calculate what will be in her account after her 100 savings.

Solution

S20 = 1050000FCFA

S40 = 4100000FCFA

b)    Sub d into eqn------(1)

20a + 190(5000) = 1050000

20a = 1050000 – 950000frs

20a = 100000

C)

50 [10000 + 495000]      =  25,250,000frs

Example III

O.k manufacturing company began production in year 2001. That year it produced 100000units. Its goal was to increase production by 25,000units each subsequent year. Determine the production.

a)       In 2004 and 2009

b)      In all up to 2009, given this goal was meet (G.C.E Question: June 2006, Paper I, Q28)

Solution

a = 100000units

d = 25000units

Tn = a + (n-1) d

a) i)    T1 = 4 years

T4 = 100000+[(4-1) 25000]

100000 + 75000 = 175,000units

ii)       T9 = 100000 + [(9-1) 25000]

100000 + 200000 = 300,000 units

b)

Example IV

A sales agent of Ekema Palm remitted eight sums of money by electronic cheques to the sales manager at the head office. These sums were in arithmetic progression with the first being 40,000frs. Determine the common difference and the eight sums paid if the last was 53000FCFA. (G.C.E Question: June 2003, Paper I, Q25 ; 2009, Paper 2,Q30)

Solution

a = 40000

d = ?

T8 = 530000FCFA

Tn = a + (n – 1) d

T8 = 40000 + (8 – 1)d = 530000

40000 + 7d = 530000

7d = 530000 – 40000

T8= a + (n-1)d

=40000 + (8-1)70000 = 40000 + 490000 = 530 000frs

Example V1

CAMEROON TEA Estate began production last year, it produced 8000 cartons of tea, it is projected that production will increase by 50 cartons each year.

a)     Determine the production in the 5th and 9th year.

b)    Find also its total production from start to when it has just increased production by 30% over the initial figure. (GCE Question: June 2003, Paper I, Q20)

Solution

a)     The first term (a) = 8000cartons,     d=50cartons

T5 =a + (n-1) d = 8000+4(50) =8200cartons

b)    Increase production by 30%

10400 = 8000 + (n-1) 50

10400 = 8000 + 50n – 50

2450 = 50n,   n=49 (this is the period in which the production will be 10400cartons)

Therefore, we have to look for the sum of the production from year 1 up to

Example vi( G.C.E Question: June 2007, Paper 1, Q28)

Mrs. Sophie saves money in a weekly thrifting scheme and this she does in Arithmetic Progression. The first savings is 500frs and the incremental saving is 200frs. How much is her account worth two weeks to the end of 2 years.?

Solution

a= 500frs, d=200frs

1 year=52weeks, 2 years = 52x2= 104weeks

n= 2 weeks to the end of year 2 i.e. = 104-2=102weeks

=51 x 21200=1,081,200FCFA

EQUIDISTANCE OF TWO TERMS

Two terms of an AP Tx and Ty are said to be equidistance from the extreme terms a and If Tx – a = Tn – Ty

Tx + Ty  = a + Tn

Example

Given the following sequence 1, 3, 5, 7, 9, 11, 13 we say that 3 and 11 are equidistance from 1 and 12 because 3 + 11 = 1 + 13

The sum of any two terms equidistance from the extreme ones is constant Tx + Ty = 2a + (n-1)d

Example

My Angel wants to ride a Jet plane which is worth 2.3 million FCFA. She embarks on a saving scheme in an AP in which she first save 150,000FCFA and increased her subsequent saving by 2,000frs each.

a)       Determine how much she will need to borrow in order to achieve her dream by the time he has saved 8 times on a monthly basis.

Solution

Total cost of Mercedes Ben = 23,000,000frs

First saving (a) = 150,000frs

Incremental saving (d) = 2,000frs

N = 8 times

Example II

Miss Smile saved 4,000frs the first week and 4,200frs next week. Given that the savings are in an AP.

a)                 Determine the saving made in the 12th week and the 16th week.

b)                Determine the total savings by the 18th week.

c)                 The ratio of the total saving in the 11th week and the 15th week.

Solution

First saving (a) = 4,000frs

Second saving = 4,200frs

Common difference (d) = 4200 – 4000 , d = 200 FRS

a)       The nth saving is Tn = a + (n-1)d

Where n = 12,    T12 = 4000 + (12 – 1) 200  = 6,200frs

n = 16,    T16 = 4000 + (16-1)2OO  = 7,000frs

b)      Total saving is

Where n = 18,

= 9 [8000 + 3400]

= 9 (11400)   =102,600frs

C)

5.5 (8000 + 2000)   = 55,000frs

7.5 (8000 + 2800  =  81,000frs

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