Question 1134278
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The 11th term is the 6th term plus the common difference 5 times, so the common difference is<br>
{{{(48-23)/5 = 25/5 = 5}}}<br>
The sum of the first 20 terms of the sequence is 20 times the average of all the terms; the average of all the terms is the average of the middle two terms -- the 10th and 11th terms.  Since the 11th term is 48 and the common difference is 5, the 10th term is 43; then the average of the terms in the sequence is (43+48)/2 = 91/2.<br>
And so the sum of the first 20 terms is<br>
{{{20(91/2) = 10*91 = 910}}}