Question 1117124
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a.<br>
The common difference d in the sequence is 9-4=5.<br>
a(2) = a(1)+1(d) = a(1)+1(5) = 4+5 = 9  [the second term is the first term, plus the common difference 1 time]
a(3) = a(1)+2(d) = a(1)+2(5) = 4+10 = 14  [the third term is the first term, plus the common difference 2 times]
...
a(n) = a(1)+(n-1)(d) = a(1)+(n-1)(5) = 4+10 = 14  [the n-th term is the first term, plus the common difference (n-1) times]<br>
Your textbook probably shows a formula something like "a(n) = a(1)+(n-1)(d)" for the n-th term in an arithmetic sequence.  But don't just memorize the formula; look at what it says and see that it is just common sense based on the definition of an arithmetic sequence.<br>
So the 25th term in your sequence is the first term, plus the common difference 24 times: 4 + 24(5) = 4+120 = 124.<br>
Answer for part a: the 25th term is 124.<br>
b.<br>
The formula you probably have for the sum of the first n terms of an arithmetic sequence is probably an ugly one; I suggest you don't try to memorize it.  Instead, think of finding the sum of the terms of an arithmetic sequence like this....<br>
(1) the sum of the terms is the number of terms, multiplied by the average of all the terms; and
(2) because the terms of an arithmetic sequence are equally spaced, the average of all the terms is just the average of the first and last terms.<br>
So to find the sum of the terms of an arithmetic sequence we only need to know the first and last terms and the number of terms.<br>
4, 7, 10, 13, ... out to 60 terms.<br>
The first term is 4; the common difference is 3; so the 60th term is the first term plus the common difference 59 times: 4+59(3) = 4+177 = 181.<br>
The average of the first and last terms -- and therefore the average of all the terms, is (4+181)/2 = 185/2.<br>
The sum of the first 60 terms is the number of terms, multiplied by the average of all the terms:<br>
{{{(60*185)/2 = 30*185 = 5550}}}<br>
Answer for part b: the sum of the first 60 terms is 5550.