Arithmetic Progression

An arithmetic progression is a sequence of numbers where the difference between the numbers in the same/Constant.

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Arithematic progression key Concepts

A.P Example 1

(a) For the sequence 2, 4, 6, 8, 10, 12, 14, 16 find the 15th term.

(b) For the sequence 14, 17, 20, 23, ... Find the,

(i) 18th term

(ii) Sum of the first 30 terms.

(c) Find the sum 24 terms of 1, 5, 9, 13, 17 ...

(d) The fifth term of an AP is 10 and the tenth term is 20, find the values of a and d, hence find the 20th term.

Solutions

(a)

The formula for finding a given term is nth = a + (n - 1)d we want to find term number 15 so we have n = 15

Find d which is the common difference

d = 4 - 2

d = 2

a = 2, n = 15 , d = 2

15th = a + (n - 1)d

15th = 2 + (15 - 1)2

15th = 2 + (14)2

15th = 2 + 28

Answer: 15th = 30

(b)

(i)

18th = a + (n - 1)d

Sequence = 14, 17, 20, 23

d = 17 - 14

d = 3

a = 14, n = 18, d = 3

18th = 14 + (18 - 1)3

18th = 14 + (17)3

18th = 14 + 51

Answer: 18th = 65

(ii)

Sum of first 30 terms

Formula

$Sum\; =\frac{\mathrm{n}}{2}(\mathrm{2a}+\mathrm{(n\; -\; 1)\; d})$

a = 14, n = 30, d = 3

$Sum\; =\frac{30}{2}(\mathrm{2(14)}+\mathrm{(30\; -\; 1)\; 3})$

$Sum\; =15(28+\mathrm{(29)\; 3})$

$Sum\; =15(28+87)$

$Sum\; =15\left(115\right)$

$Sum\; =1725$

Answer $Sum30\; =1725$

(c)

Sequence is 1, 5, 9, 13, 17

Sum of the first 24 terms

d = 5 - 1

d = 4

a = 1, n = 24, d = 4

$Sum\; =\frac{\mathrm{n}}{2}(\mathrm{2a}+\mathrm{(n\; -\; 1)\; d})$

$Sum\; =\frac{24}{2}(\mathrm{2(1)}+\mathrm{(24\; -\; 1)\; 4})$

$Sum\; =12(2+\mathrm{(23)\; 4})$

$Sum\; =12(2+92)$

$Sum\; =12\left(94\right)$

$Sum\; =1128$

Answer: $Sum24\; =1128$

(d)

The formula for term of an AP is nth = a + (n - 1)d

Fifth term is 10

Meaning term number 5 is 10, so n = 5, then term number 5 is 10. So we have;

T5 = a + (5 - 1)d

T5 = a + (4)d

a + 4d = 10 ....(i)

The tenth term is 20

Meaning term number 10 is 20, so n = 10, then term number 10 is 20. So we have;

T10 = a + (10 - 1)d

T10 = a + (9)d

a + 9d = 20 ....(ii)

Solve equation i and ii simultaneously

a + 4d = 10 ....(i)

a + 9d = 20 ....(ii)

a = 10 - 4d ....(i)

substitute the value a into the equaction (ii)

a + 9d = 20 ....(ii)

10 - 4d + 9d = 20 ....(ii)

- 4d + 9d = 20 - 10

5d = 10

$\frac{\mathrm{5d}}{5}=\frac{\mathrm{10}}{5}$

Answer: $d=2$

substitute d into the equation below

a = 10 - 4d ....(i)

a = 10 - 4(2)

a = 10 - 8

Answer: a = 2

Find the 20th term

nth = a + (n - 1)d

a = 2 , n = 20 , d = 2

nth = a + (n - 1)d

20th = 2 + (20 - 1)2

20th = 2 + (19)2

20th = 2 + 38

Answer: 20th = 40

A.P Example 2

For the sequence 2, 5, 8, 11, 14 ... find

(i) An expression for the nth term

(ii) The 9th term using an expression in (i)

(iii) The 100th term using an expression in (i)

(iv) An expression for the sum of the first nth terms

(v) The sum of the first 40 terms using an expression in (i)

Solutions

(i)

Formula nth = a + (n - 1)d

d = 5 - 2

d = 3

a = 2, n = ? d = 3

nth = 2 + (n - 1)3

nth = 2 + 3n - 3

nth = 3n + 2 - 3

Answer: nth = 3n - 1

(ii)

9th term

nth = 3n - 1

9th = 3(9) - 1

9th = 27 - 1

Answer: 9th = 26

(iii)

100th term

nth = 3n - 1

100th = 3(100) - 1

100th = 300 - 1

Answer: 100th = 299

(iv)

Formula $Sum\; =\frac{\mathrm{n}}{2}(\mathrm{2a}+\mathrm{(n\; -\; 1)\; d})$

a = 2 , d = 3, n = ?

$Sum\; =\frac{\mathrm{n}}{2}(\mathrm{2(2)}+\mathrm{(n\; -\; 1)\; 3})$

$Sum\; =\frac{\mathrm{n}}{2}(4+\mathrm{3n\; -\; 3})$

$Sum\; =\frac{\mathrm{n}}{2}\left(\mathrm{3n\; +\; 4\; -\; 3}\right)$

$Sum\; =\frac{\mathrm{n}}{2}\left(\mathrm{3n\; +\; 1}\right)$

Answer: $Sth\; =\frac{{\mathrm{3n}}^2}{2}+\frac{\mathrm{n}}{2}$

(iv)

$Sth\; =\frac{{\mathrm{3n}}^2}{2}+\frac{\mathrm{n}}{2}$

n = 40th

$S40\; =\frac{{\mathrm{3(40)}}^2}{2}+\frac{40}{2}$

$S40\; =\frac{\mathrm{3(1600)}}{2}+20$

$S40\; =\frac{\mathrm{4800}}{2}+20$

$S40\; =\mathrm{2400}+20$

Answer: $S40\; =2420$

A.P Example 3