Answer: 4th term=-5/6 Sn=-5/6 Step-by-step explanation: a=1/6,d=1/4-1/6=-1/2 4th term =a+3d =1/6-3/2 =1-6/6 =-5/6 Sn= a+(n-1)d = 1/6+(4-1)×-1/2 = 1/6+(3×-1/2) = 1/6-3/2 = 1-6/6 =-5/6 Reply
Answer: [tex]\huge\boxed{\dfrac{5}{12},\ \dfrac{1}{2},\ \dfrac{7}{12},\ \dfrac{2}{3}}\\\boxed{S_n=\dfrac{n^2+3n}{24}}[/tex] Step-by-step explanation: [tex]a_1=\dfrac{1}{6},\ a_2=\dfrac{1}{4},\ a_3=\dfrac{1}{3}\\\\a_2-a_1=\dfrac{1}{4}-\dfrac{1}{6}=\dfrac{1\cdot3}{4\cdot3}-\dfrac{1\cdot2}{6\cdot2}=\dfrac{3}{12}-\dfrac{2}{12}=\dfrac{1}{12}\\\\a_3-a_2=\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{1\cdot4}{3\cdot4}-\dfrac{1\cdot3}{4\cdot3}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}[/tex] It’s an arithmetic sequence where: [tex]a_1=\dfrac{1}{6},\ d=\dfrac{1}{12}[/tex] The explicit formula: [tex]a_n=a_1+(n-1)d\\\\a_n=\dfrac{1}{6}+(n-1)\left(\dfrac{1}{12}\right)=\dfrac{1}{6}+\dfrac{1}{12}n-\dfrac{1}{12}=\dfrac{1\cdot2}{6\cdot2}-\dfrac{1}{12}+\dfrac{1}{12}n=\dfrac{2}{12}-\dfrac{1}{12}+\dfrac{1}{12}n\\\\\boxed{a_n=\dfrac{1}{12}+\dfrac{1}{12}n=\dfrac{1+n}{12}}[/tex] Calculate next 4 terms: substitute n = 4, n = 5, n = 6, n = 7 [tex]a_4=\dfrac{1+4}{12}=\dfrac{5}{12}\\\\a_5=\dfrac{1+5}{12}=\dfrac{6}{12}=\dfrac{1}{2}\\\\a_6=\dfrac{1+6}{2}=\dfrac{7}{12}\\\\a_7=\dfrac{1+7}{12}=\dfrac{8}{12}=\dfrac{2}{3}[/tex] [tex]S_n=\dfrac{[2a_1+(n-1)d]\cdot n}{2}\\\\S_n=\dfrac{\left[2\cdot\frac{1}{6}+(n-1)\left(\frac{1}{12}\right)\right]\cdot n}{2}=\dfrac{\left(\frac{1}{3}+\frac{1}{12}n-\frac{1}{12}\right)\cdot n}{2}=\dfrac{\left(\frac{1\cdot4}{3\cdot4}-\frac{1}{12}+\frac{1}{12}n\right)\cdot n}{2}\\\\=\dfrac{\left(\frac{4}{12}-\frac{1}{12}+\frac{1}{12}n\right)\cdot n}{2}=\dfrac{\left(\frac{3}{12}+\frac{1}{12}n\right)\cdot n}{2}=\dfrac{1}{2}\cdot\left(\dfrac{3}{12}+\dfrac{1}{12}n\right)\cdot n[/tex] [tex]\boxed{S_n=\dfrac{3}{24}n+\dfrac{1}{24}n^2=\dfrac{n^2+3n}{24}}[/tex] Reply
Answer:
4th term=-5/6
Sn=-5/6
Step-by-step explanation:
a=1/6,d=1/4-1/6=-1/2
4th term =a+3d
=1/6-3/2
=1-6/6
=-5/6
Sn= a+(n-1)d
= 1/6+(4-1)×-1/2
= 1/6+(3×-1/2)
= 1/6-3/2
= 1-6/6 =-5/6
Answer:
[tex]\huge\boxed{\dfrac{5}{12},\ \dfrac{1}{2},\ \dfrac{7}{12},\ \dfrac{2}{3}}\\\boxed{S_n=\dfrac{n^2+3n}{24}}[/tex]
Step-by-step explanation:
[tex]a_1=\dfrac{1}{6},\ a_2=\dfrac{1}{4},\ a_3=\dfrac{1}{3}\\\\a_2-a_1=\dfrac{1}{4}-\dfrac{1}{6}=\dfrac{1\cdot3}{4\cdot3}-\dfrac{1\cdot2}{6\cdot2}=\dfrac{3}{12}-\dfrac{2}{12}=\dfrac{1}{12}\\\\a_3-a_2=\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{1\cdot4}{3\cdot4}-\dfrac{1\cdot3}{4\cdot3}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}[/tex]
It’s an arithmetic sequence where:
[tex]a_1=\dfrac{1}{6},\ d=\dfrac{1}{12}[/tex]
The explicit formula:
[tex]a_n=a_1+(n-1)d\\\\a_n=\dfrac{1}{6}+(n-1)\left(\dfrac{1}{12}\right)=\dfrac{1}{6}+\dfrac{1}{12}n-\dfrac{1}{12}=\dfrac{1\cdot2}{6\cdot2}-\dfrac{1}{12}+\dfrac{1}{12}n=\dfrac{2}{12}-\dfrac{1}{12}+\dfrac{1}{12}n\\\\\boxed{a_n=\dfrac{1}{12}+\dfrac{1}{12}n=\dfrac{1+n}{12}}[/tex]
Calculate next 4 terms:
substitute n = 4, n = 5, n = 6, n = 7
[tex]a_4=\dfrac{1+4}{12}=\dfrac{5}{12}\\\\a_5=\dfrac{1+5}{12}=\dfrac{6}{12}=\dfrac{1}{2}\\\\a_6=\dfrac{1+6}{2}=\dfrac{7}{12}\\\\a_7=\dfrac{1+7}{12}=\dfrac{8}{12}=\dfrac{2}{3}[/tex]
[tex]S_n=\dfrac{[2a_1+(n-1)d]\cdot n}{2}\\\\S_n=\dfrac{\left[2\cdot\frac{1}{6}+(n-1)\left(\frac{1}{12}\right)\right]\cdot n}{2}=\dfrac{\left(\frac{1}{3}+\frac{1}{12}n-\frac{1}{12}\right)\cdot n}{2}=\dfrac{\left(\frac{1\cdot4}{3\cdot4}-\frac{1}{12}+\frac{1}{12}n\right)\cdot n}{2}\\\\=\dfrac{\left(\frac{4}{12}-\frac{1}{12}+\frac{1}{12}n\right)\cdot n}{2}=\dfrac{\left(\frac{3}{12}+\frac{1}{12}n\right)\cdot n}{2}=\dfrac{1}{2}\cdot\left(\dfrac{3}{12}+\dfrac{1}{12}n\right)\cdot n[/tex]
[tex]\boxed{S_n=\dfrac{3}{24}n+\dfrac{1}{24}n^2=\dfrac{n^2+3n}{24}}[/tex]