There is a common difference of 2. For the nth term a₁=7, a₂=9, a₃=11, a₄=13, a₅=15 where n goes from 1 to 5. Subtract 1 from each term: 6, 8, 10, 12, 14. Now halve each: 3, 4, 5, 6, 7, which is 1+2, 2+2, 3+2, 4+2, 5+2.

So, by reversing the operations, a[n] (meaning a sub n)=2(n+2)+1=2n+5. (The reverse of ½ is 2 and the reverse of subtract is add.) Check: if n=1 a₁=2×1+5=7; n=2, a₂=2×2+5=9.

Therefore a[n]=2n+5.

- All categories
- Pre-Algebra Answers 12,323
- Algebra 1 Answers 25,303
- Algebra 2 Answers 10,448
- Geometry Answers 5,189
- Trigonometry Answers 2,642
- Calculus Answers 6,028
- Statistics Answers 3,026
- Word Problem Answers 10,140
- Other Math Topics 6,678

81,777 questions

86,081 answers

2,216 comments

69,578 users