Question

An=17+2n s12-? s5-13-?

Answer 1

A1=19

A5=17+2*5=27

A12=17+2*12=41

s12=(a1+a12)12/2=360

S5-13=(a1+a5)5/2-13=102

Answer 2

$a_n=17+2n\\a_1=17+2\cdot1=19\\\ a_4=17+2\cdot4=25\\a_{12}=17+2\cdot12=41\\a_{13}=17+2\cdot13=43\\ \\ S_{12}=\frac{(a_1+a_{12})\cdot12}{2} =(19+41)\cdot6= 360\\\\ S_{4}=\frac{(a_1+a_{4})\cdot4}{2} =(19+25)\cdot2=44\cdot2=88\\ \\S_{13}=S_{12}+a_{13}=360+43=403\\ \\ S_{5-13}=S_{13}-S_4=403-88=315\\ \\ OTBET:S_{12}=360;S_{5-13}=315$

Answer 3

To find the sum of an arithmetic sequence, you can use the formula for the nth term of an arithmetic sequence:

a_n = a_1 + (n - 1) * d

Where:

• a_n is the nth term
• a_1 is the first term
• n is the term number
• d is the common difference between consecutive terms

In your case, the first term (a_1) is 17, and the common difference (d) is 2. Now, you want to find the sum of the first 12 terms (S_12) and the sum of the first 5 terms (S_5).

1. Sum of the first 12 terms (S_12): Using the formula for the sum of an arithmetic series:

   S_n = (n/2) * [2a_1 + (n - 1)d]

   Plug in the values: S_12 = (12/2) * [2(17) + (12 - 1)(2)] S_12 = 6 * [34 + 22] S_12 = 6 * 56 S_12 = 336

So, the sum of the first 12 terms (S_12) is 336.

2. Sum of the first 5 terms (S_5): Using the same formula:

   S_5 = (5/2) * [2(17) + (5 - 1)(2)] S_5 = (5/2) * [34 + 8] S_5 = (5/2) * 42 S_5 = 5 * 21 S_5 = 105

So, the sum of the first 5 terms (S_5) is 105.