Arithmetic Progression (AP) [Mathematics JAMB Tutorial]

Today on our JAMB UTME Free Tutorial, will treat mathematics and our topic is Arithmetic Progression (AP); SEQUENCE AND SERIES.

As we said in the introduction, we will be teaching you ARITHMETIC PROGRESS (AP). Since you can’t talk about AP without starting mentioning SEQUENCE and SERIES, we will start by explaining those terms.

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In a case where numbers are presented in a given array or in an orderly manner, such set of numbers is regarded as SEQUENCE.

EXAMPLE:

• 1,2,3,4…
• 2,4,6,8,10…
• -3,-2,-1…

Also, when the summation of a series is taken, the process is called SERIES

EXAMPLE:

• 1+2+3+4+…
• 2+4+6+8+10+…

ARITHMETIC PROGRESSION (AP)

AP is a form of sequence in which each term or preceding number is obtained by the addition of a certain number called the COMMON DIFFERENCE.

EXAMPLE 2:

Examine the pattern of the AP below

• 7,9,11,13,15…

SOLUTION

Nth TERM OF AN AP

The nth term of an arithmetic progression is usually donated as “a” and its common difference by “d” while the last term of an AP may be denoted as “n”

We can understand this better using example 2 above and we can have that:

In Summary, the general term of an AP is given as:

• T_n = a +  (n  –  1) d

EXAMPLE 3

Find the 4^th term of a AP whose first term is 2 and common difference is 0.5

SOLUTION:

T_n = a + (n – 1)d………………(1)

Here, n=4, a=2, d= 0.5

Substituting (1) above we have

T₄ = 2 + (4 – 1) 0.5

= 2 +(3)0.5

= 2+1.5

 = 3.5

SUM OF AN AP

The sum of an AP refers to the addition of all the terms mentioned or required in a given AP. It is usually donated by S_n . The formulae is given by

Sn = n/2 {2a + (n – 1)d}…………….(2) this is used when a, d, and n is given

Sn = n/2 (a + l) ……………………….(3) this is used when the first term and the last term is given.

EXAMPLE 4:

The first and last term of a linear sequence (A.P) are -12 and 40 respectively if the sum of the sequence is 196, find;

1. The number of the terms
2. The common difference
3. The 12^th term

Solution 4:

(1) P last and first term is given by;

Sn =n/2(a+L)

196=n/2(-12+40)

196 x 2 = n(28)

196 x 2/28 = n

14 = n

(2) To find d, we apply the general sum formula for AP

Sn = n/2{2a + (n – 1) d}

S₁₄ → 196 = 14/2 { 2(-12) + (14 – 1)d}

196 = 7(-24 + 13d)

196/7 = -24 + 13d

28 + 24 = 13d

52/13 =d

4 = d

(3) Find the 12th term

Tn = a + (n-1)d

T₁₂ = -12 +( 12-1)4

= -12 + (11)4

= -12 +44

T₁₂ =  32

EXERCISES ON AP

1. If -8, m, n, 19 are in AP. Find (m,n)
   (a) 1, (b) 10 2, 10 (c) 3, 13 (d) 4, 16
2. The 6^th term of an AP is 20 and the 11^th term is -5. Find the sum of the first three terms
   (a) 45 (b) 85 (c) 120 (d) 150
3. Find the nth term of sequence 4, 10, 16, …
   a. 2(3n – 1) b. 2(2 + 3^n-1) c. 2ⁿ + 2 d. 2(3n + 2)

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