Question

in an arithmetic sequence the sum of the 2nd and 6th term is 32. given that the sum of the first six terms is 120. Find the tenth term of the sequence

Answer 1

a₁₀ = - 32

Explanation:

the nth term of an arithmetic sequence is

•
`a_(n)` = a₁ + d(n - 1)

a₁ is the first term, d the common difference , n the term number

given

a₂ + a₆ = 32, that is

a₁ + d + a₁ + 5d = 32 ( simplify left side )

2a₁ + 6d = 32 → (1)

the sum of n terms of an arithmetic sequence is

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

given S₆ = 120 , then

120 =
`(6)/(2)` [ 2a₁ + 5d ]

120 = 3(2a₁ + 5d ) ← divide both sides by 3

40 = 2a₁ + 5d , that is

2a₁ + 5d = 40 → (2)

using the 2 equations to solve simultaneously for a₁ and d

subtract (2) from (1) term by term to eliminate a₁

(2a₁ - 2a₁) + (6d - 5d) = 32 - 40

0 + d = - 8

d = - 8

substitute d = - 8 into (1) and solve for a₁

2a₁ + 6(- 8) = 32

2a₁ - 48 = 32 ( add 48 to both sides )

2a₁ = 80 ( divide both sides by 2 )

a₁ = 40

Then

a₁₀ = a₁ + d(n - 1) substitute a₁ = 40 and d = - 8

a₁₀ = 40 + (- 8 × 9 ) = 40 + (- 72) = 40 - 72 = - 32