contestada

The first three terms of an arithmetic sequence are u1=0.5
u
1
=
0.5
, u2=1.8
u
2
=
1.8
, u3=3.1
u
3
=
3.1
.

(a) Find the common difference.

[2]

(b) Find the 30
30
th term of the sequence.

[2]

(c) Find the sum of the first 30
30
terms.

Respuesta :

mdejazulhaque1972

Answer:

1)Common difference=1.3 as 3.1-1.8=1.3 also 1.8-0.5=1.3

2)Method to take out Progression

a+(n-1)d=tn

0.5+(30-1)1.3=tn

0.5+37.7=38.2

3)580.5

Ver imagen mdejazulhaque1972
Ver imagen mdejazulhaque1972

Answer:

see explanation

Step-by-step explanation:

(a)

The common difference d = u₂ - u₁ = 1.8 - 0.5 = 1.3

(b)

The nth term of an arithmetic sequence is

[tex]u_{n}[/tex] = u₁ + (n - 1)d

Here u₁ = 0.5 and d = 1.3 , then

u₃₀ = 0.5 + (29 × 1.3) = 0.5 + 37.7 = 38.2

(c)

The sum to n terms of an arithmetic sequence is

[tex]S_{n}[/tex] = [tex]\frac{n}{2}[/tex] [ 2u₁ + (n - 1)d ] , then

[tex]S_{30}[/tex] = [tex]\frac{30}{2}[/tex] [ (2 × 0.5) + (29 × 1.3) ]

     = 15(1 + 37.7)

    = 15 × 38.7

    = 580.5

Otras preguntas