the 4th term of an ap is equal to 3 times the first term and the 7th term exceeds twice the third term by 1 find the first term and the common difference

We're given

a₄ = 3a₁

a₇ = 2a₃ + 1

Since these numbers form an arithmetic progression, we have for some fixed constant d

a₂ = a₁ + d

a₃ = a₂ + d = a₁ + 2d

a₄ = a₃ + d = a₁ + 3d

and so on, up to

a₇ = a₁ + 6d

Rewrite the given equations in terms of a₁, then solve for a₁ and d :

a₁ + 3d = 3a₁

a₁ + 6d = 2 (a₁ + 2d) + 1

2a₁ - 3d = 0

a₁ - 2d = -1

(2a₁ - 3d) - 2 (a₁ - 2d) = 0 - 2 (-1)

2a₁ - 3d - 2a₁ + 4d = 0 + 2

d = 2

a₁ + 3d = 3a₁

3 (2) = 2a₁

a₁ = 3

y=−9/2x+10

step-by-step explanation:

let's solve for y.

0−9x−2y=−20

step 1: add 9x to both sides.

−9x−2y+9x=−20+9x

−2y=9x−20

step 2: divide both sides by -2.

−2y

−2

=

9x−20

−2

y=

−9

2

x+10

answer: for each 3 girls there is 2 boys . 3: 2

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