<PRE><B>Question 1173732</B>
&lt;br&gt;
The sum of 8 terms is 160, so the average of the 8 terms is 160/8=20.&lt;br&gt;
The average of the 8 terms is the average of the first term, a, and the 8th term, a+7d:&lt;br&gt;
{{{(a+(a+7d))/2 = 20}}}
{{{2a+7d = 40}}} (1)&lt;br&gt;
The sum of 20 terms is 880, so the average of the 20 terms is 880/20 = 44.&lt;br&gt;
The average of the 20 terms is the average of the first term, a, and the 20th term, a+19d:&lt;br&gt;
{{{(a+(a+19d))/2 = 44}}}
{{{2a+19d = 88}}} (2)&lt;br&gt;
Comparing (1) and (2), we see 12d=48, so the common difference d is 4.&lt;br&gt;
Then either (1) or (2) tells us that the first term is 6.&lt;br&gt;
ANSWERS:&lt;br&gt;
43rd term: a+42d = 6+42(4) = 6+168 = 174&lt;br&gt;
Sum of (first) 12 terms: 12 times the average of the first and 12th:&lt;br&gt;
{{{12((a+(a+11d))/2) = 12((2a+11d)/2) = 12((12+44)/2) = 6(56) = 336}}}&lt;br&gt;
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