<PRE><B>Question 844510</B>
Each term of sequence is determined by adding half to the preceding term.The sum of the first twenty five terms of the sequence equals the square of the twenty fifth term. How can I calculate the possible value of the first term? 
&lt;pre&gt;
Substitute d=0.5 and n=25 in the formula for the nth term 
of an arithmetic sequence.

 a&lt;sub&gt;n&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt; + (n-1)d
a&lt;sub&gt;25&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt; + (25-1)(.5)
a&lt;sub&gt;25&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt; + (24)(.5)
a&lt;sub&gt;25&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt; + 12

Substitute the same values d=0.5 and n=25 in the formula
for the sum of the first n terms of an arithmetic sequence.
 
 S&lt;sub&gt;n&lt;/sub&gt; = {{{n/2}}}[2a&lt;sub&gt;1&lt;/sub&gt; + (n-1)d] 
S&lt;sub&gt;25&lt;/sub&gt; = {{{25/2}}}[2a&lt;sub&gt;1&lt;/sub&gt; + (25-1)(.5)]
S&lt;sub&gt;25&lt;/sub&gt; = {{{25/2}}}[2a&lt;sub&gt;1&lt;/sub&gt; + (24)(.5)]
S&lt;sub&gt;25&lt;/sub&gt; = {{{25/2}}}[2a&lt;sub&gt;1&lt;/sub&gt; + 12]
S&lt;sub&gt;25&lt;/sub&gt; = 25a&lt;sub&gt;1&lt;/sub&gt; + 150
&lt;/pre&gt;
The sum of the first twenty five terms of the sequence equals the square of the twenty fifth term
&lt;pre&gt;
So 

       S&lt;sub&gt;25&lt;/sub&gt; = (a&lt;sub&gt;25&lt;/sub&gt;)²

25a&lt;sub&gt;1&lt;/sub&gt; + 150 = (a&lt;sub&gt;1&lt;/sub&gt; + 12)²

25a&lt;sub&gt;1&lt;/sub&gt; + 150 = (a&lt;sub&gt;1&lt;/sub&gt;)² + 24a&lt;sub&gt;1&lt;/sub&gt; + 144

         0 = (a&lt;sub&gt;1&lt;/sub&gt;)² - a&lt;sub&gt;1&lt;/sub&gt; - 6

         0 = (a&lt;sub&gt;1&lt;/sub&gt; + 2)(a&lt;sub&gt;1&lt;/sub&gt; - 3)

      a&lt;sub&gt;1&lt;/sub&gt; + 2 = 0;  a&lt;sub&gt;1&lt;/sub&gt; - 3 = 0
         
         a&lt;sub&gt;1&lt;/sub&gt; = -2;  a&lt;sub&gt;1&lt;/sub&gt; = 3

So there are two solutions for the first term, -2 and 3

Checking both by computer, gave these lists of 25 terms and 
the sums all the way to the 25th term:

 1.  term = -2 , sum = -2       1.  term = 3 , sum = 3
 2.  term = -1.5 , sum = -3.5   2.  term = 3.5 , sum = 6.5
 3.  term = -1 , sum = -4.5     3.  term = 4 , sum = 10.5
 4.  term = -0.5 , sum = -5     4.  term = 4.5 , sum = 15
 5.  term = 0 , sum = -5        5.  term = 5 , sum = 20
 6.  term = 0.5 , sum = -4.5    6.  term = 5.5 , sum = 25.5
 7.  term = 1 , sum = -3.5      7.  term = 6 , sum = 31.5
 8.  term = 1.5 , sum = -2      8.  term = 6.5 , sum = 38
 9.  term = 2 , sum = 0         9.  term = 7 , sum = 45
10.  term = 2.5 , sum = 2.5    10.  term = 7.5 , sum = 52.5
11.  term = 3 , sum = 5.5      11.  term = 8 , sum = 60.5
12.  term = 3.5 , sum = 9      12.  term = 8.5 , sum = 69
13.  term = 4 , sum = 13       13.  term = 9 , sum = 78
14.  term = 4.5 , sum = 17.5   14.  term = 9.5 , sum = 87.5
15.  term = 5 , sum = 22.5     15.  term = 10 , sum = 97.5
16.  term = 5.5 , sum = 28     16.  term = 10.5 , sum = 108
17.  term = 6 , sum = 34       17.  term = 11 , sum = 119
18.  term = 6.5 , sum = 40.5   18.  term = 11.5 , sum = 130.5
19.  term = 7 , sum = 47.5     19.  term = 12 , sum = 142.5
20.  term = 7.5 , sum = 55     20.  term = 12.5 , sum = 155
21.  term = 8 , sum = 63       21.  term = 13 , sum = 168
22.  term = 8.5 , sum = 71.5   22.  term = 13.5 , sum = 181.5
23.  term = 9 , sum = 80.5     23.  term = 14 , sum = 195.5
24.  term = 9.5 , sum = 90     24.  term = 14.5 , sum = 210
25.  term = 10 , sum = 100     25.  term = 15 , sum = 225
          
Using a&lt;sub&gt;1&lt;/sub&gt; = -2, a&lt;sub&gt;25&lt;/sub&gt; = 10, S&lt;sub&gt;25&lt;/sub&gt; = 100 which is 10².
Using a&lt;sub&gt;1&lt;/sub&gt; = 3, a&lt;sub&gt;25&lt;/sub&gt; = 15, S&lt;sub&gt;25&lt;/sub&gt; = 225 which is 15².
 
Edwin&lt;/pre&gt;</PRE>