Bunuel wrote:
a1, a2, a3, ..., an
In the sequence above, each term after the first term is equal to the preceding term plus the constant c. If \(a_1+a_3+a_5=27\), what is the value of \(a_2 + a_4\) ?
A. 5
B. 9
C. 13
D. 18
E. 22
Took me quite some time to figure out how to get to an answer, since I was not able to calculate a definite value for \(a_1\)
\(a_1+a_3+a_5=27\) \(-->\) \(a_3=a_1+2c\) and \(a_5=a_1+4c\)
Rewrite the equation: \(a_1+a_1+2c+a_1+4c=27\) \(-->\) \(a_1+2c=9\) At this points I was quite stuck... But I then assumed integer values for \(c\)
\(c=0\) \(-->\) \(a_1=9\) \(-->\) \(a_2 + a_4=9+9=18\)
\(c=1\) \(-->\) \(a_1=7\) \(-->\) \(a_2 + a_4=8+10=18\)
\(c=2\) \(-->\) \(a_1=5\) \(-->\) \(a_2 + a_4=7+11=18\)
\(c=3\) \(-->\) \(a_1=3\) \(-->\) \(a_2 + a_4=6+12=18\)
\(c=4\) \(-->\) \(a_1=1\) \(-->\) \(a_2 + a_4=5+13=18\)
As we can see any value of \(a_1\) will result in \(a_2 + a_4=18\), hence D.
Bunuel is there any faster way to solve this question? I was quite stuck thinking i was doing something wrong, because I could not solve for a value of \(a_1\)
_________________