Exercise 202 Page 558

a Consider the given finite arithmetic series.

We can see that the first term is t(1)= 3, the common difference is d= 0.3, and the last term is t(n)= 63. Let's recall the explicit formula of an arithmetic sequence, which will help us find the number of terms in the given series. t(n) = t(1) + (n-1)d First, we will find the explicit formula for the sequence describing the series! Let's substitute d= 0.3 and t(1)= 3 in this formula and simplify.

t(n) = t(1) + (n-1)d

SubstituteII

d= 0.3, t(1)= 3

t(n) = 3 + (n-1)* 0.3

Now we will find the number of terms in our series by substituting t(n) = 63 and solving for n. Let's go!

t(n)=3+(n-1)0.3

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Solve for n

Substitute

t(n)= 63

63 = 3+(n-1)* 0.3

SubEqn

LHS-3=RHS-3

60 = (n-1)* 0.3

DivEqn

.LHS /0.3.=.RHS /0.3.

200 = n-1

AddEqn

LHS+1=RHS+1

201 = n

RearrangeEqn

Rearrange equation

n = 201

The last term is the 201^(th) term of the series. This means that the series has 201 terms.

b Now we want to find the sum of the series. Let's recall the formula for the sum S_n of a finite arithmetic series.

S(n)=n(t(1)+t(n))/2 In Part A, we found that the given series has n = 201 terms. Let's substitute n=201, t(1)= 3, and t(n)= 63 into the formula and simplify.

S(n)=n(t(1)+t(n))/2

SubstituteValues

Substitute values

S(201)=201( 3+ 63)/2

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Simplify right-hand side

AddTerms

Add terms

S(201) = 201(66)/2

CalcQuot

Calculate quotient

S(201) = 201(33)