Question

Please answer this question​

Answer 1

`a_n=3n-2`

Explanation:

General form of an arithmetic sequence:
`a_n=a+(n-1)d`

where:

• 
  `a_n` is the nth term
• 
  `a` is the first term
• 
  `d` is the common difference between terms

Create expressions for the 4th and 6th terms:

`\implies a_4=a+(4-1)d=a+3d`

`\implies a_6=a+(6-1)d=a+5d`

The ratio of the 4th term to the 6th term is 5:8, therefore:

`\implies (a_4)/(a_6)=(5)/(8)`

`\implies (a+3d)/(a+5d)=(5)/(8)`

`\implies 8(a+3d)=5(a+5d)`

`\implies 8a+24d=5a+25d`

`\implies 8a-5a=25d-24d`

`\implies 3a=d \quad \leftarrow \textsf{Equation 1}`

Sum of the first n terms of an arithmetic series:

`S_n=(n)/(2)[2a+(n-1)d]`

The sum of the first 7 terms of an arithmetic progression is 70:

`\implies S_7=70`

`\implies (7)/(2)[2a+(7-1)d]=70`

`\implies 2a+6d=20`

`\implies a+3d=10 \quad \leftarrow \textsf{Equation 2}`

Substitute Equation 1 into Equation 2 and solve for
`a`:

`\implies a+3(3a)=10`

`\implies a+9a=10`

`\implies 10a=10`

`\implies a=1`

Substitute found value of
`a` into Equation 1 and solve for
`d`:

`\implies 3(1)=d`

`\implies d=3`

Finally, substitute found values of
`a` and
`d` into the general form of the arithmetic sequence:

`\implies a_n=1+(n-1)3`

`\implies a_n=1+3n-3`

`\implies a_n=3n-2`