Bunuel wrote:
\(t_1\), \(t_2\), \(t_3\), ..., \(t_n\), ...
In the sequence above, each term after the first term is equal to the preceding term plus the constant k. If \(t_1+t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\) ?
A. 8
B. 12
C. 24
D. 32
E. 72
Kudos for a correct solution.
According to the question, t2=t1+k, t3=t2+k=t1+k+k So, t3=t1+2k, t4=t1+3k, t5=t1+4k.. That's the pattern.
So, t1+t3+t5+t7
=t1+t1+2k+t1+4k+t1+6k
=4t+12k which is 32 given
At this point, u can take any value of t and k that satisfies the equation 4t+12k=32. (The (t,k) pairs could be (2,2) or (5,1).
Now, we are asked to find t2+t4+t6, which can be written as 3t+k+3k+5k, which in turn is equal to 3t+9k.
Taking either pair of values of (t,k), we get 3t+9k=6+18=24.
So answer is B
Alternatively, you could see that as t1+t3+t5+t7=32, and as 1+3+5+7=16, so the total, to be 32, each term has to be double, i.e. 2+6+10+14=32. Now these terms are not consecutive meaning there is one term between any two consecutive of these terms. So, t2+t4+t6 would be 4+8+12, which is 24.
Hope it helps.