How to solve a sequence when you don't know the first term?

Question

How to figure out the first five terms term in an arithmetic ande geometric sequence and solve?

1. arithmetic, a sub2 = -2, d= -5 (Do I subtract -5 from -2 to get first term?

2. geometric, a sub 2 = -2, a sub 3 = -3 (How do I determine the common ration?)

Many thanks for your help!

Answer 1

1) Yup. a_1 - 5 = a_2, so a_1 must be 3.
2) What does the common ratio actually mean? Its any term divided by the term before it. So in this case, the common ratio is just (-3)/(-2) = 1.5. Now, how would you get the first term? To go from the first term to the second you'd multiply by 1.5; thus to go from the second to the first, divide by 1.5: -2/1.5 = -4/3 is the first term.

Answer 2

You are correct about the first part, as long as you know that subtracting a negative number is is like adding the positive.

In a geometric sum, one term is multiplied by another to get the next term. In this case, -2 times something = -3. That can be written as -2x=-3. Now it's just solve for x. x=-3/-2 or 1.5. That being know we can now work backwards.

Something times 1.5 = -2.
1.5x=-2 x=-2/1.5 x=-1.333 with a line over the threes or -4/3.

The first term is -4/3.

Answer 3

an = a1 + (n - 1)d
a2 = a1 + (2 - 1)(-5)
-2 = a1 + 1(-5)
-2 = a1 - 5
a1 = 3

Using this you get

an = 3 - 5(n - 1)

a1 = 3 - 5(1 - 1) = 3 - 5(0) = 3 - 0 = 3
a2 = 3 - 5(2 - 1) = 3 - 5(1) = 3 - 5 = -2
a3 = 3 - 5(3 - 1) = 3 - 5(2) = 3 - 10 = -7
a4 = 3 - 5(4 - 1) = 3 - 5(3) = 3 - 15 = -12
a5 = 3 - 5(5 - 1) = 3 - 5(4) = 3 - 20 = -17

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an = a1 * r^(n - 1)

a2 = a1 * r^(2 - 1)
-2 = a1 * r^1
a1 = (-2/r)

a3 = a1 * r^(3 - 1)
-3 = a1 * r^2
a1 = -3/(r^2)

Now just set the 2 equal to each other.

(-2/r) = (-3/(r^2))

cross multiply

-2r^2 = -3r

divid both sides by r

-2r = -3

divide both sides by -2

r = (3/2)

Using that you get

Now plug this in for r into the equation a1 = (-2/r)

a1 = -2/(3/2)
a1 = (-2/1)/(3/2)
a1 = (-2/1)*(2/3)
a1 = (-4/3)

and now you have

an = (-4/3) * (3/2)^(n - 1)

a1 = (-4/3) * (3/2)^(1 - 1) = (-4/3)(3/2)^0 = (-4/3) * 1 = (-4/3)
a2 = (-4/3) * (3/2)^(2 - 1) = (-4/3)(3/2)^1 = (-4/3)(3/2) = -2
a3 = (-4/3) * (3/2)^(3 - 1) = (-4/3)(3/2)^2 = (-4/3)(9/4) = -3
a4 = (-4/3) * (3/2)^(4 - 1) = (-4/3)(3/2)^3 = (-4/3)(27/8) = (-9/2)
a5 = (-4/3) * (3/2)^(5 - 1) = (-4/3)(3/2)^4 = (-4/3)(81/16) = (-27/4)

If you like, you can put the answers in Mixed Fractions

Go to www.ltcconline.net/greenl/Courses/103B/seqSeries/ARITSEQ.HTM
and
www.mathguide.com/lessons/SequenceGeometric.html

to see where i got my help from.

Answer 4

it really is an mathematics progression so sum is (first + very last) x (0.5 the shape of words) yet first we favor to comprehend the way many words there are sixty seven = 18 + 7(n - a million) => 40 9 = 7(n - a million) => n = 8 Sn = (18 + sixty seven) x 4 = 80 5 x 4 = 340 i do not comprehend what 'locate the sum of the first 4 words contained in the sequence of 2n-a million' ability, although.

Answer 5

set up the equation first: a sub n = d(n-1)+a1
you know a sub 2 is -2 and d is -5 so sub -2 for a sub n and -5 for d
we have: -2= -5(n-1)+a1
given that -2 is in second position so 2 for n
-2= -5(2-1)+a1
-2= -5(1)+a1
-2= -5+a1
a1=3

Answer 6

are you only given those two terms? and what is d= -5?