The 3rd and 7th terms of an arithmetic progression are 6and 30 respectively determine the common difference, first term,10th term.​

d = 6 , a₁ = - 6 and a₁₀ = 48

Step-by-step explanation:

The nth term of an arithmetic progression is

= a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₃ = 6 and a₇ = 30 , then

a₁ + 2d = 6 → (1)

a₁ + 6d = 30 → (2)

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

4d = 24 ( divide both sides by 4 )

d = 6

Substitute d = 6 into (1)

a₁ + 2(6) = 6

a₁ + 12 = 6 ( subtract 12 from both sides )

a₁ = - 6

Then

a₁₀ = - 6 + (9 × 6) = - 6 + 54 = 48

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