Question 1161060
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Let the first term be a(1) and the common difference be d.<br>
(1) a(10) is 7 times a(3):<br>
{{{a[1]+9d = 7(a[1]+2d)}}}
{{{6a[1]+5d = 0}}}<br>
(2) The sum of the first 30 terms is 2460:<br>
{{{30(a[1]+a[30])/2 = 2460}}}
{{{30((a[1])+(a[1]+29d)) = 4920}}}
{{{2a[1]+29d = 164}}}<br>
Solve the pair of equations from (1) and (2):<br>
{{{6a[1]+87d = 492}}}
{{{6a[1]+5d = 0}}}
{{{82d = 492}}}
{{{d = 6}}}<br>
The common difference d is 6; use it to find a(1):<br>
{{{6a[1]+5(6) = 0}}}
{{{a[1] = -30/6 = -5}}}<br>
The first term is -5; the common difference is 6.<br>
The 6th term is the first term, plus the common difference 5 times:<br>
t(6) = -5+5(6) = -5+30 = 25<br>