andrewantonov07
Answered

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an arithmetic sequence has a 10th term of 15 and 14th term of 35

show the equation (y=mx+b) of this graph equals an = -30+(n-1)5​

Sagot :

kimcpi

Answer:

an=5n−35

Step-by-step explanation:

The question says to show the equation in (y=mx+b) form.

So, we convert -30+(n-1)5​ into (y=mx+b) form.

we get rid of the parenthesis by multiplying 5 with (n-1)

n×5 = 5n

-1×5=-5

So we are left with -30+5n-5

We solve for like terms, -30-5

so we are left with 5n-35.

an=5n-35

Now we double check if this equation is valid, the question states that the 10th term is 15 and the 14th term is 35. We plug both in, replace of n

10th term is 15

an = 5(10)-35

an = 50-35

an=15 ✔ (that checks out)

14th term is 35

an = 5(14)-35

an = 70-35

an=35 ✔ (that also checks out)

THEREFORE, the equation in (y=mx+b) form would be

an=5n−35

mhanifa

Answer:

  • See below

Step-by-step explanation:

Given an AP with:

  • a₁₀ = 15
  • a₁₄ = 35

General form for nth term:

  • aₙ = a₁ + (n - 1)d

Apply this to the given terms:

  • a₁ + 9d = 15
  • a₁ + 13d = 35

Solve the system above by elimination, subtract the equations:

  • 13d - 9d = 35 - 15
  • 4d = 20
  • d = 5

Find the first term:

  • a₁ + 9*5 = 15
  • a₁ = 15 - 45
  • a₁ = - 30

Now, substitute the values of the first term and common difference into general equation:

  • aₙ = a₁ + (n - 1)d
  • aₙ = - 30 + (n - 1)*5
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