Question

please answer this question​

Answer 1

Answer:

$a_n=3n-2$

Step-by-step 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:

$\implies \dfrac{a_4}{a_6}=\dfrac{5}{8}$

$\implies \dfrac{a+3d}{a+5d}=\dfrac{5}{8}$

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

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

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

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

Sum of the first n terms of an arithmetic series:

$S_n=\dfrac{n}{2}[2a+(n-1)d]$

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

$\implies S_7=70$

$\implies \dfrac{7}{2}[2a+(7-1)d]=70$

$\implies 2a+6d=20$

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

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:

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

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

$\implies a_n=3n-2$