Question

the sum of the terms of the arithmetic sequence. 1, 7, 13, 19, . . ., 109 n = 19

Answer 1

you add 6 every time, and thus
109 = 108+1=1 + (6 × 18) so 109 is the 18th term.

So, you can say now,

S(n) = ½ (1+109) × 18

Answer 2

Need more help ?

Answer 3

my bad 109 is the 19th term

Answer 4

because you skip 18 times

Answer 5

S(n) = ½ (1+109) × 19 (not 18, but 19) Like this.

Answer 6

\[t_{n}=a+(n-1)d,109=1+(n-1)6,109-1=6(n-1),n-1=\frac{ 108 }{ 6 }=18\]
n=18+1=19
\[s _{n}=\frac{ n \left( a+t _{n} \right) }{ 2 },s _{19}=\frac{ 19\left( 1+109 \right) }{ 2 }=\frac{ 19\times110 }{ 2 }=?\]