Question 1182882
<br>
The sum of the first n terms of an arithmetic progression with first term a and constant difference d is<br>
{{{S(n)=(n(2a+(n-1)d))/2}}}<br>
We will use that formula twice in solving this problem.<br>
The sum of the first ten terms is<br>
{{{S(10) = (10(2a+9d))/2}}}<br>
The sum of the 20th, 21st, and 22nd terms is three times the 21st term:<br>
{{{3(a+20d)}}}<br>
The problem then tells us<br>
{{{(10(2a+9d)/2)=3(a+20d)}}}
{{{10a+45d=3a+60d}}}
{{{7a=15d}}}<br>
The sequence contains only positive integers; and the first term a is less than 20.  Since 7 and 15 have no common factor, the only solution that meets the requirements is a=15 and d=7.  So the progression is<br>
15, 22, 29, 36, ...<br>
The problem says to find the number of terms n for which the sum is 960:<br>
{{{S(n) = (n(30+(n-1)7))/2 = 960}}}<br>
{{{30n+7n^2-7n = 1920}}}
{{{7n^2+23n-1920 = 0}}}<br>
Since that won't factor easily, use the quadratic formula to find n=15.<br>
ANSWER: 15 terms of the sequence will have a sum of 960.<br>
CHECK:
1st term 15
10th term 15+9(7)=78
21st term 15+20(7)=155<br>
sum of first 10 terms: {{{(10(15+78))/2=5(93)=465}}}
sum of 20th, 21st, and 22nd terms: {{{3(155)=465}}}<br>