Question

Does anyone know how to do this?? Please explain

Answer 1

see explanation

Explanation:

in an arithmetic sequence there is a common difference d between consecutive terms , then

(a)

a₂ - a₁ = a₃ - a₂ , that is

- k + 10 - (2k + 1) = k - 1 - (- k + 10) ← distribute parenthesis on both sides

- k + 10 - 2k - 1 = k - 1 + k - 10

- 3k + 9 = 2k - 11 ( add 3k to both sides )

9 = 5k - 11 ( add 11 to both sides )

20 = 5k ( divide both sides by 5 )

4 = k

(b)

the first 3 terms are then

a₁ = 2k + 1 = 2(4) + 1 = 8 + 1 = 9

a₂ = - k + 10 = - 4 + 10 = 6

a₃ = k - 1 = 4 - 1 = 3

the first 3 terms are 9, 6, 3

(c)

d = a₂ - a₁ = 6 - 9 = - 3

(d)

the sum to n terms of an arithmetic sequence is

`S_(n)` =
`(n)/(2)` [ 2a₁ + (n - 1)d ]

where a₁ is the first term and d the common difference

here a₁ = 9 and d = - 3 , then

S₂₀ =
`(20)/(2)` [ (2 × 9 ) + (19 × - 3) ]

= 10( 18 - 57)

= 10 × - 39

= - 390