
Concept explainers
To find:The 25th partialsum of given arithmetic sequence.

Answer to Problem 66E
The 25th partial sum of the given arithmetic sequence is 1850 .
Explanation of Solution
Given information:
An arithmetic sequence is given as 2,8,14,20,...
Concept used:
An arithmetic sequence of n terms, has the form a1,a2,a3,...,an and should have a common difference between each two consecutive terms.
That is
a2−a1=a3−a2=a4−a3=...=an−an−1
Common difference can be defined by d .
a2−a1=a3−a2=a4−a3=...=an−an−1=d
nth term of the arithmetic sequence has the form
an=a1+(n−1)d
Where a1 is the first term of the sequence, and d is the common difference.
Sum of an arithmetic finite sequence has the form
Sn=n2(a1+an)
Here, n is number of terms, a1 is the first term of the sequence, and an is the last term of the sequence,
Calculation:
Consider the given sequence.
2,8,14,20,...
Now, first term of the finite sequence is a1=2 .
Common difference is
d=a2−a1=8−2=6
25th term of the sequence can be found as
. an=a1+(n−1)da25=2+(25−1)6=2+144=146
Number of term for the required sumis indicated as 25.
So, the 25th partial sum is calculated as
Sn=n2(a1+an)S25=252(2+146)=25×74=1850
Thus, the 25th partial sum of the given arithmetic sequence is 1850 .
Chapter 8 Solutions
Precalculus with Limits: A Graphing Approach
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