Question 1147514
<br>
Given: 1st term 6; 5th term 18<br>
The difference is 12; the 5th term is 4 terms after the first.  So the common difference is 12/4 = 3<br>
With a common difference of 3 and a first term of 6, the formula for the n-th term of the sequence is 3n+3.<br>
The sum of n terms of an arithmetic progression is<br>
(number of terms) times (average of all the terms)<br>
which in an arithmetic progression is<br>
(number of terms) times (average of first and last terms)<br>
We have expressions for all those numbers:
number of terms: n
first term: 6
last (n-th) term: 3n+3<br>
We want to find the number n that makes the sum 162:<br>
{{{n((6+(3n+3))/2) = 162}}}
{{{n((3n+9)/2) = 162}}}
{{{n(3n+9) = 324}}}
{{{n(n+3) = 108}}}<br>
Formal algebra or trial and error shows n=9 (the other algebraic solution n=-12 doesn't make sense in the context of the problem).<br>
ANSWER: The number of terms to have a sum of 162 is 9.<br>
CHECK:
1st term: 6
9th term: 6+8*3 = 6+24 = 30
Sum of 9 terms: 9((6+30)/2) = 9(36/2) = 9(18) = 162