The first two terms of a sequence are a1 =4 and a2= -2. Let a3 be the third term when the sequence is arithmetic and let b3 be the third term when the sequence is geometric. Find a3+b3

a3 + b3 = -7

Step-by-step explanation:

When the sequence is A.P

a1 =4 and a2= -2

Common difference,d = a2 - a1

= -2 - 4

d = -6

a3 = third term

a3 = a1 + 2d

= 4 + 2(-6)

= 4 - 12

= -8

a3 = -8

When the sequence is geometric

a1 =4 and a2= -2

Common ratio, r = a2/a1

= -2/4

r = -1/2

Or

r = -0.5

b3 = third term

b3 = ar²

= 4 * (-1/2)²

= 4 * 1/4

= 4/4

= 1

b3 = 1

a3 + b3 = -8 + 1

= -7

a3 + b3 = -7

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.