In the sequence above, each term after the first term is equal to the

Ans: C
Solution: according to given condition
t2=t1+k
t3=t2+k = t1+2k
same way t4=t1+3k
t5=t1+4k
t6=t1+5k
t7=t1+6k

now by putting the values in given equation
t1+t3+t5+t7=32
t1+3k=8

required value
t2+t4+t6 = 3(t1+3k)= 3*8= 24

Ans: C

OFFICIAL SOLUTION:

According to the stem:
\(t_2=t_1+k\);
\(t_3=t_2+k=t_1+2k\);
\(t_4=t_3+k=t_1+3k\);
...
\(t_n=t_1+(n-1)k\);

Since \(t_1+t_3+t_5+t_7=32\), then \(t_1+(t_1+2k)+(t_1+4k)+(t_1+6k)=32\). So, \(t_1+3k=8\)

We need to find the of \(t_2+t_4+t_6=(t_1+k)+(t_1+3k)+(t_1+5k)=3(t_1+3k)\). From above we know that \(t_1+3k=8\), thus \(3(t_1+3k)=3*8=24\).

Answer: C.

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.

t1t1, t2t2, t3t3, ..., tntn, ...
In the sequence above, each term after the first term is equal to the preceding term plus the constant k. If t1+t3+t5+t7=32t1+t3+t5+t7=32, what is the value of t2+t4+t6t2+t4+t6 ?

A. 8
B. 12
C. 24
D. 32
E. 72


t1+t3+t5+t7=32
➡4*t1+12k=32
➡t1+3k=8
t1=2
k=2
t2+t4+t6=4+8+12=24
C

Another method is to note that the average of the two sequences would be the same.

The average of \(t_1\), \(t_3\), \(t_5\) and \(t_7\) will be the middle of \(t_3\) and \(t_5\) which is \(t_4\). (All terms in an AP are equidistant from the term before it and the term after it.)
The average of \(t_2\), \(t_4\), \(t_6\) is obviously \(t_4\).

\(t_1+t_3+t_5+t_7=4 * t_4 = 32\)

\(t_4 = 8\)

Then \(t_2+t_4+t_6 = 3 * t_4 = 24\)

Answer (C)

What about the box symbol that is in front of t3, t5 and t4? How are we saying that t3 = box t3?

That was some bug. Edited. Thank you.

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