Find the 30th term of the arithmetic sequence with a first term of 4 and a comman difference of 8

Mathematics

1 answer:

Arte-miy333 [17]3 years ago

3 0

Answer:

236

Step-by-step explanation:

The n th term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 4 and d = 8, thus

$a_{30}$ = 4 + (29 × 8) = 4 + 232 = 236

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Answer the Blanks:

vfiekz [6]

Answer:

1. 90°

2. 90°

3. 40°

4. 45°

5. 45°

Step-by-step explanation:

Given: ΔABC,

CD⊥AB,

m∠A=50°,

m∠ACB=85°

Solution:

1. ∠ADC is a right ange, because CD⊥AB, so

$m\angle ADC=90^{\circ}$

2. ∠CDB is a right ange, because CD⊥AB, so

$m\angle CDB=90^{\circ}$

3. Consider triangle ACD. The sum of the measures of all interior angles is always 180°, so

$m\angle A+m\angle ADC+m\angle ACD=180^{\circ}\\ \\50^{\circ}+90^{\circ}+m\angle ACD=180^{\circ}\\ \\m\angle ACD=180^{\circ}-50^{\circ}-90^{\circ}=40^{\circ}$

4. By Angle Addition Postulate,

$m\angle ACD+m\angle BCD=m\angle ACB\\ \\40^{\circ}+m\angle BCD=85^{\circ}\\ \\m\angle BCD=85^{\circ}-40^{\circ}=45^{\circ}$

5. Consider triangle ABC. The sum of the measures of all interior angles is always 180°, so

$m\angle A+m\angle ACB+m\angle B=180^{\circ}\\ \\50^{\circ}+85^{\circ}+m\angle B=180^{\circ}\\ \\m\angle B=180^{\circ}-50^{\circ}-85^{\circ}=45^{\circ}$