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Marizza181 [45]
3 years ago
13

The 9th term of an arithmetic series is 5/2

Mathematics
1 answer:
Verdich [7]3 years ago
8 0

Answer:

- 8050

Step-by-step explanation:

The n th term of an arithmetic sequence is

a_{n} = a + (n - 1)d

where a is the first term and d the common difference.

We require to find both a and d

Given the 9 th term is 2.5 , then

a + 8d = 2.5 → (1)

Given the sum of the second and fifth term is 27, then

a + d + a + 4d = 27, that is

2a + 5d = 27 → (2)

Multiply (1) by - 2 and add to (2) to eliminate a

- 2a - 16d = - 5 → (3)

Add (2) and (3) term by term

- 11d = 22 ( divide both sides by - 11 )

d = - 2

Substitute d = - 2 into (1) and solve for a

a - 16 = 2.5 ( add 16 to both sides )

a = 18.5

The sum to n terms of an arithmetic sequence is

S_{n} = \frac{n}{2}[ 2a + (n - 1)d ], thus

S_{100} = 50 [ (2 × 18.5) + (99 × - 2) ] = 50(37 - 198) = 50(- 161) = - 8050

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1. Angles ADC and CDB are supplementary, thus

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Since m∠ADC=115°, you have that m∠CDB=180°-115°=65°.

2. Triangle BCD is isosceles triangle, because it has two congruent sides CB and CD. The base of this triangle is segment BD. Angles that are adjacent to the base of isosceles triangle are congruent, then

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The sum of the measures of interior angles of triangle is 180°, therefore,

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