In an arithmetic sequence u2 = 3 and u4 = 7. Find u15.

Step-by-step solution

1. 1

   Get $d$ from the gap between the two known terms. $u_4-u_2=(4-2)d=2d\quad\Longrightarrow\quad 2d=7-3=4,\ d=2$

2. 2

   Back out the first term. $u_2=u_1+d$, so $u_1=3-2=1$

3. 3

   Apply the nth-term rule for $n=15$. $u_{15}=u_1+14d=1+14\cdot 2=29$

4. 4

   Or jump straight from $u_4$. $u_{15}=u_4+(15-4)d=7+11\cdot 2=29$, the same value reached without ever computing $u_1$.

5. 5

   Reject the near-miss. The distractor $31$ is $u_{16}$ (one step too far). The answer is D, $u_{15}=29$.