Find the sum of the first 16 terms of an arithmetic sequence where a4=7 and a7= 16. (I need the procedure y Example) please

Mathematics

1 answer:

Luden [163]1 year ago

6 0

Answer:

S₁₆ = 328

Step-by-step explanation:

the nth term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

given a₄ = 7 and a₇ = 16 , then

a₁ + 3d = 7 → (1)

a₁ + 6d = 16 → (2)

subtract (1) from (2) term by term to eliminate a₁

0 + 3d = 9

3d = 9 ( divide both sides by 3 )

d = 3

substitute d = 3 into (1) and solve for a₁

a₁ + 3(3) = 7

a₁ + 9 = 7 ( subtract 9 from both sides )

a₁ = - 2

the sum to n terms of an arithmetic sequence is

$S_{n}$ = $\frac{n}{2}$ [ 2a₁ + (n - 1)d ]

with a₁ = - 2 and d = 3 , then

S₁₆ = $\frac{16}{2}$ [ (2 × - 2) + (15 × 3) ]

= 8(- 4 + 45)

= 8 × 41

= 328

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Classify each sequence as arithmetic, geometric or neither

Karolina [17]

Answer:

Geometric sequence = ⁵/9, ⁵/3, 5, 15, 45. . .

Arithmetic sequence = 16, 11, 6, 1, -4 . . .

Step-by-step explanation:

A geometric sequence is a sequence of numbers, in which every number in the sequence after the first term is multiplied by a common ratio, which gives us the next term. The common ratio is the ratio of a term to the term before it.

On the other hand, an arithmetic sequence is a sequence of number have a common difference between each consecutive terms. This common difference is gotten by subtracting a term from the term before it.

Now let's examine each sequence given:

==>16, 11, 6, 1, -4 . . .

11 - 16 = -5

6 - 11 = -5

1 - 6 = -5

-4 - 1 = -5

This is an arithmetic sequence. The commons difference "-5" is the same

==>⁵/9, ⁵/3, 5, 15, 45. . .

⁵/3 ÷ ⁵/9 = ⁵/3 × ⁹/5 = ⁹/3 = 3

5 ÷ ⁵/3 = ⁵ × ⅗ = 3

15 ÷ 5 = 3

45 ÷ 15 = 3