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>
Answer:
C
Step-by-step explanation:
The last one.
Hope this helped!
WHERES THE QUESTION IT WONT SHOW UP ON MY SCREEN POST IT AGAIN
Answer:
2.14
Step-by-step explanation:
2.14+3.82= 5.96
5.96-3.82= 2.14
So, your answer would be 2.14
