Question

Find the next 4 terms of the sequence 1
/6,1/4,1/3
also find Sn.​

Answer 1

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 2

Answer:

$\huge\boxed{\dfrac{5}{12},\ \dfrac{1}{2},\ \dfrac{7}{12},\ \dfrac{2}{3}}\\\boxed{S_n=\dfrac{n^2+3n}{24}}$

Step-by-step explanation:

$a_1=\dfrac{1}{6},\ a_2=\dfrac{1}{4},\ a_3=\dfrac{1}{3}\\\\a_2-a_1=\dfrac{1}{4}-\dfrac{1}{6}=\dfrac{1\cdot3}{4\cdot3}-\dfrac{1\cdot2}{6\cdot2}=\dfrac{3}{12}-\dfrac{2}{12}=\dfrac{1}{12}\\\\a_3-a_2=\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{1\cdot4}{3\cdot4}-\dfrac{1\cdot3}{4\cdot3}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}$

It's an arithmetic sequence where:

$a_1=\dfrac{1}{6},\ d=\dfrac{1}{12}$

The explicit formula:

$a_n=a_1+(n-1)d\\\\a_n=\dfrac{1}{6}+(n-1)\left(\dfrac{1}{12}\right)=\dfrac{1}{6}+\dfrac{1}{12}n-\dfrac{1}{12}=\dfrac{1\cdot2}{6\cdot2}-\dfrac{1}{12}+\dfrac{1}{12}n=\dfrac{2}{12}-\dfrac{1}{12}+\dfrac{1}{12}n\\\\\boxed{a_n=\dfrac{1}{12}+\dfrac{1}{12}n=\dfrac{1+n}{12}}$

Calculate next 4 terms:

substitute n = 4, n = 5, n = 6, n = 7

$a_4=\dfrac{1+4}{12}=\dfrac{5}{12}\\\\a_5=\dfrac{1+5}{12}=\dfrac{6}{12}=\dfrac{1}{2}\\\\a_6=\dfrac{1+6}{2}=\dfrac{7}{12}\\\\a_7=\dfrac{1+7}{12}=\dfrac{8}{12}=\dfrac{2}{3}$

$S_n=\dfrac{[2a_1+(n-1)d]\cdot n}{2}\\\\S_n=\dfrac{\left[2\cdot\frac{1}{6}+(n-1)\left(\frac{1}{12}\right)\right]\cdot n}{2}=\dfrac{\left(\frac{1}{3}+\frac{1}{12}n-\frac{1}{12}\right)\cdot n}{2}=\dfrac{\left(\frac{1\cdot4}{3\cdot4}-\frac{1}{12}+\frac{1}{12}n\right)\cdot n}{2}\\\\=\dfrac{\left(\frac{4}{12}-\frac{1}{12}+\frac{1}{12}n\right)\cdot n}{2}=\dfrac{\left(\frac{3}{12}+\frac{1}{12}n\right)\cdot n}{2}=\dfrac{1}{2}\cdot\left(\dfrac{3}{12}+\dfrac{1}{12}n\right)\cdot n$

$\boxed{S_n=\dfrac{3}{24}n+\dfrac{1}{24}n^2=\dfrac{n^2+3n}{24}}$