What is the sum of the 12th to 20th terms (inclusive) of the arithmetic sequence 7,12,17,22,...?

Question

The solution to this question told that you should do S(20) - S(12). But I did S(20) - S(11). Here's my reasoning:

We are asked to find the sum of the 12th to 20th terms (inclusive), which I think means U12 + U13 + ... + U20.

S(20) = U1 + ... + U11 + U12 + U13 + U14 + U15 + U16 + U17 + U18 + U19 + U20

S(11) = U1 + ... + U11

Thus, to get U12 + ... + U20, you do S(20) - S(11).

I'm not sure why the answer says S(20) - S(12). Am I not interpreting the word "inclusive" properly?

Answer 1

Without computing each of these sums , you have a direct formula for that: there are $9$ terms in this sum, so, denoting $u_n$ the $n$-th term, the sum is $$ S=9\,\frac{u_{12}+u_{20}}2.$$

Answer 2

You have the famous formula $$\sum_{1}^{n} = \dfrac{n(n+1)}2$$

The sum from the question is actually $$2n+5\sum_{1}^{n} = 2n+5\dfrac{n(n+1)}2$$

From the $12^{th}$ to the $20^{th}$

$$2\cdot 20+5\dfrac{20(20+1)}2 - 2\cdot 11+5\dfrac{11(11+1)}2$$