the 3rd term of an A. P is 6 and the 16th term is 32 find the common difference and the sum of the first 20 term

2 and 420

Step-by-step explanation:

The n th term of an AP is

= a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₃ = 6 and a₁₆ = 32 , then

a₁ + 2d = 6 → (1)

a₁ + 15d = 32 → (2)

Subtract (1) from (2) term by term

13d = 26 ( divide both sides by 13 )

d = 2

Substitute d = 2 into (1) for corresponding value of a₁

a₁ + 2(2) = 6

a₁ + 4 = 6 ( subtract 4 from both sides )

a₁ = 2

The sum of n terms of an AP is

= [ 2a₁ + (n - 1)d ] , thus

= [(2 × 2) + (19 × 2) ]

     = 10(4 + 38) = 10 × 42 = 420

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18

answer:

step-by-step explanation: