Probability Maths Questions (1 Viewer)

[Schniz]

I have no idea how to even begin these questions....



Q, the sum of the first three terms of an arithmetic sequnece is 24 and the sum of the next three terms is 51. find the first term and the common difference of this sequence.

Q. If x= (2ab) / (a+b) show that 1/a, 1/x, 1/b are in arithmetic progression.

Q. Find the sum of all the positive integers less that 100, which are not divisible by 6.

Q. Find the sum of all the integers between 100 and 200 which are multiples of 9.

 

[webby234]

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

S₃ = 24
S₆ = 24 + 51 = 75

24 = 3/2[2a + 2d]
16 = 2a + 2d
25 = 2a + 5d
3d = 9
d = 3
a = 5

 

[webby234]

2) If x= (2ab) / (a+b) show that 1/a, 1/x, 1/b are in arithmetic progression.

If true 1/x - 1/a = 1/b - 1/x

(a+b)/2ab - 1/a = 1/b - (a+b)/2ab
(a+b - 2b) = 2a - (a+b)
a-b = a-b

 

[webby234]

3) Integers that are multiples of 6 are 6,12,18...96

Arithmetic series with
a = 6
d = 6

96 is the final term
96 = a + d(n-1)
96 = 6 + 6(n-1)
15 = n- 1
n = 16

S₁₆ = 8[12 + 6 x 15]
= 816

Sum of integers from one to 100
= 1 + 100 + 2 + 99...50 + 51
= 50 x 101
= 5050

So sum of integers that are not multiples of 6
= 5050 - 816
= 4234

 

[Schniz]

thanks heaps buddy

 

[webby234]

From 100 to 200 multiples of 9 are

108, 117 ... 198

AS with a = 108 d = 9

198 = 108 + 9(n-1)
10 = n - 1
n = 11

S₁₁ = 11/2[216 + 9 x 10]
= 1683

 

[Schniz]

ur a champ, thx so much