Question

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

Answer 1

d=6

a=-6

Explanation:

use the formula for the nth term which is

Tn=a+(n-1)d..you will have to create two equations then solve them as a simultaneous equation

T3=6 and T7=30

T3=a+(3-1)d

6=a+2d........... first equation

T7=a+(7-1)d

30=a+6d.......... second equation

then solve them as a simultaneous equation

a+2d=6

a+6d=30

-4d/-4=-24/-4

d=6

a+2d=6

a+2(6)=6

a=6-12

a=-6

I hope this helps

Answer 2

d = 6 , a₁ = - 6 and a₁₀ = 48

Explanation:

The nth term of an arithmetic progression is

`a_(n)` = 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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