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4 years ago

The 4th term of an arithmetic sequence is 12 and the 8th term is 36. Find the 17th term of the sequence.

Mathematics

1 answer:

Blizzard [7]4 years ago

6 0

Answer: 90

Step-by-step explanation:

The formula for calculating the nth term of a sequence is given as :

$t_{n}$ = a + ( n - d )

Where a is the first term

d is the common difference and

n is the number of terms

This means that the 4th term of an arithmetic sequence will have the formula :

$t_{4}$ = a + 3d

And the 4th term has been given to be , 12 ,substituting into the formula we have

12 = a + 3d .............................. equation 1

Also substituting for the 8th term , we have

36 = a + 7d .............................. equation 2

Combining the two equations , we have

a + 3d = 12 ................... equation 1

a + 7d = 36 ------------ equation 2

Solving the system of linear equation by substitution method , make a the subject of formula from equation 1 , that is

a = 12 - 3d ................... equation 3

substitute a = 12 - 3d into equation 2 , equation 2 then becomes

12 - 3d + 7d = 36

12 + 4d = 36

subtract 12 from both sides

4d = 36 - 12

4d = 24

divide through by 4

d = 6

substitute d = 6 into equation 3 to find the value of a, we have

a = 12 - 3d

a = 12 - 3 ( 6)

a = 12 - 18

a = -6

Therefore , the 17th term of the sequence will be :

$t_{17}$ = a + 16d

$t_{17}$ = -6 + 16 (6)

$t_{17}$ = -6 + 96

$t_{17}$ = 90

Therefore : the 17th term of the sequence is 90

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kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$