Sum of the first 3 terms of and ap is 33 and then the sum of the next 4 is 128

Answer 1

S₇ = 161

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Answer 2

D is the common difference

X1 is the first term

X1 + D is the second term

X1 + 2D is the third term

Their sum is 33...

3*X1 + 3D = 33

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The 4th term is x1 + 3d

the 5th term is x1 + 4d

the 6th term is x1 + 5d

the 7th term is x1 + 6d

Their sum is 4*x1 + 18d = 128

3*X1 + 3d = 33

4*x1 +18d = 128

Multiplies top equation by -4 and bottom equation by 3:

-12*x1 + -12*d = -132

12*X1 + 54*d = 384

42*d = 252

d=6

then

3*X1 + 3(6) = 33

3*x1 + 18 = 33

3*x1 = 15

x1 = 5

5,11,17,23,29,35,41