d = 3 , a₁₂ = 40 and S
= 7775
In an arithmetic sequence the nth term and sum to n terms are
<h3>• a

= a₁ + (n-1)d</h3><h3>• S

=

[2a + (n-1)d]</h3><h3>
where d is the common difference</h3><h3>a₆ = a₁ + 5d = 22 ⇒ 7 + 5d = 22 ⇒ 5d = 15 ⇔ d = 3</h3><h3>a₁₂ = 7 + 11d = 7 +( 11× 3) = 7 + 33 = 40</h3><h3>S₁₀₀ =

[(2×7) +(99×3)</h3><h3> = 25(14 + 297) = 25(311)= 7775</h3>
The common ratio of the given geometric sequence is the number that is multiplied to the first term in order to get the second term. Consequently, this is also the number multiplied to the second term to get the third term. This cycle goes on and on until a certain term is acquired. In this item, the common ratio r is,
r = t⁵/t⁸ = t²/t⁵
The answer, r = t⁻³.
The next three terms are,
n₄ = (t²)(t⁻³) = t⁻¹
n₅ = (t⁻¹)(t⁻³) = t⁻⁴
n₆ = (t⁻⁴)(t⁻³) = t⁻⁷
The answers for the next three terms are as reflected above as n₄, n₅, and n₆, respectively.
The widths of the bars are equal to ensure that numbers aren't represented improperly. If a bar was wider than another that was the same height, it could be misinterpreted as being larger.
Answer:
14x
Step-by-step explanation:
please mark this answer as the brainlest