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3 years ago

The 6th term of an arithmetic progression is 35 and the 13th term is 77. find the 1st term

Mathematics

1 answer:

GenaCL600 [577]3 years ago

8 0

Answer:

Step-by-step explanation:

The 6th term of an arithmetic progression is 35 and the 13th term is 77. find the 1st term

We know we have an arithmetic sequence.

a_n = (n - 1)*d + a_1

given:

a_6 = 35 = (6-1)*d + a_1

a_13 = (13 -1)*d + a_1 = 77

find a_1

we have 35 = 5d + a_1

and 77 = 12d + a_1

2 equations in 2 unknowns.

We can solve this.

77 = 12d + a_1

35 = 5d + a_1

minus the 2 equations from each other.

77 - 35 = 12d - 5d + a_1 - a_1

42 = 7d

d = 6

Find a_1

35 = 5*6 + a_1

35 = 30 + a_1

a_1 = 35 -30 = 5

a_1 = 5

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Step-by-step explanation:

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3 years ago

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3 years ago

The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct

lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

$y=\dfrac{x}{1+x}$

Point $P\left(1,\dfrac{1}{2}\right)$ lies on the curve.

Point $Q\left(x,\dfrac{x}{1+x}\right)$ is an arbitrary point on the curve.

The slope of the secant line PQ is

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]$