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:
-3/6 -2/6 -1/6 0/6 1/6 2/6
Step-by-step explanation:
Sum of angles in a triangle = 180°.
Question 3:
We have 90° + 61° + (14x + 1)° = 180°.
=> 14x° + 152° = 180°
=> 14x° = 28°
=> x = 2.
Question 4:
We have 70° + 55° + (4x + 3)° = 180°.
=> 4x° + 128° = 180°
=> 4x° = 52°
=> x = 13.
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