Answer:
98
You have to use prime factor decompisition. I hope this helps
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>
⇒

is a improper fraction.

as a mixed number would be 4

⇒10 x 4 = 40
⇒40 + 9 = 49

= 4
⇒ANSWER = 4
Step-by-step explanation:
this is correct answer thank you
Answer:
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