INTRODUCTION TO LINEAR AND EXPONENTIAL SEQUENCE

 

Arithmetic progression AP

* Formulae derivation

a + (a + d) + (a + 2d)+........+(l - d) + l

Where

a = is the first term

d = is the common difference

l = last term

        l = a + (n - 1)d

Normal equation is given by↖

But If we ask to find the sum of an AP with number of terms n, is given by

S = 1/2(n) [a + ι] and for finding the last term.

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

 example from the above formulae

Examples of Complete square method

Example 1 

 Findthe sum of the first 12 terms of the arithmetic progression

5 + 8 + 11 +.....

Solution

 a = 5, d = to find the common difference we must subtract the first term from the second  term, therefore 

T₂ - T₁ = 8 - 5 = 3 = d

where n = 12

Using this formulae

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

We have : 

1/2(12)[2(5) + (12 -1)3

6[10 + (11)3]

6[10 + (33)]

6(43) = 258             good!

 Example2

The first and the last term of an AP are 79 and -5 and it's sum is 814

a) the number of the terms in the AP

b) the common difference between them.

Solution to a)

S = 1/2(n) (a + ι)

a = 79, d = ?, ι = -5, S = 814

814 = 1/2(n)(79 - 5)

814 = n/2 (74)

814 × 2 = n(74)

1628/74 = n

22 = n

b)

 T₂ - T₁ = d but we dont have them in series sequence.

a +(n - 1)d = ι

79 + (22 - 1)d = -5

79 + 21d = -5

21d = -5 - 79

d = -84/21 = -4

d = -4.     Take a break with this challenge box

       Welcome To GP

 Geometric Progression GP

a + ar + ar² + ....... +arⁿ⁻¹

a = is the first term

r = is thecommon ratio

n = number of terms 

arⁿ⁻¹

Sn = a(rⁿ - 1)/r-1   only if   |r| > 1

Sn = a(1 - rⁿ)/1-r    only if  |r| < 1

S∞ = a/1-r sum of infinity

Example 1

Find the number of terms in following GP

2 + 4 + 8 +.......+512

a = 2, r = 8/4 =2

2(2)ⁿ⁻¹ = 512

4ⁿ⁻¹ = 512

4(n - 1) = 512

4n - 4 = 512

4n = 512 + 4

4n = 516

n = 516/4

n = 129

Example 2

A GP has its third term as 10 and six term as 80 

find  a, r , and S₆

T₃= 10 = ar² ..,....... Eq1

T₆ = 80 = ar⁵ .......... Eq2

 Now let's take the ratio

ar⁵/ar² = 80/10

Now using the law of Indices we have.

r³ = 8 therefore 2³ = 8

r = 2.  n = 3

ar³⁻¹ = ar² = 10

ar² = 10

a(2)² = 10

a4 =10

a = 10/4 =5/2

S₆= 5/2 (2⁶ - 1)/2 -1

S₆ = 5/2 (64 - 1)/1

S₆ = 5/2 (63)

S₆ = 315/2  

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