Arithmetic Sequences(Please help - Don't understand)

Question

1. Find the the 25^th term of an arithmetic sequence s whose 16^th term is 11/2 and whose 41^st term is 43/3
2. If s is an arithmetic sequence (with associated series S ) such that the 28th term = -101 and the 4th term = 4 , find the 11th and 20th term.
3. Let  s be an arithmetic sequence whose difference is 6 and whose 101st term is 596. One of the terms of its associated series S is 230 . Which term?
4. Find real numbers x and y and  such that 16,x ,y ,and 54 are four consecutive terms (in the order given) of a geometric sequence.

Answer 1

The terms in an arithmetic sequence are given by:

a_n = a₁+ (n-1)d.

Part 1 gives you two terms,a₁₆ and a₄₁. With these 2 values you set up a pair of simultaneous linear equations with variable a₁ and d. Once you solve the set for a and d, you can compute a₂₅.

_{Part 2 works the same way.}

_{For part 3 you need the sum of n terms of an arithmetic sequence which is an+n(n-1)d/2. Note that what is given is a partial sum, i.e. 230. You first need to find the first term.}

For part 4 you need to know that the terms of a geometric sequence are arⁿ where a is the first term, r is the ratio and n starts at 0.

Answer 2

In these problems more time is spent on fractions than on learning how the formulas work!

a₄₁ = a₁₆ + 24d

43/3 = 11/2 + 24d

43/3 - 11/2 = 24d

53/6 = 24d

53/144 = d

a₂₅ = a₁₆ + 8(53/144)

a₂₅ 11/2 + 53/18

a₂₅ = 77/9

2) and 3) are done with the same process

4)

54 = 16(r)³

54/16 = r³

27/8 = r³

3/2 = 6r

Can you determine x and y?