Question

Q60
Please help me as soon as possible.

Answer 1

16.875

Explanation:

`\boxed{\begin{minipage}{5.5 cm}\underline{Sum of an infinite geometric series}\\\\$S_(\infty)=(a_1)/(1-r)$,\;\;for\;$|r| < 1$\\\\where:\\\phantom{ww}$\bullet$ $a_1$ is the first term. \\ \phantom{ww}$\bullet$ $r$ is the common ratio.\\\end{minipage}}`

Given:

• Sum of infinite geometric series:
  `S_(\infty) = 120`
• Common ratio: r = 0.75

To find the value of the third term, we first need to find the value of the first term (a₁) by substituting the given values of
`S_(\infty)` and r into the sum of an infinite geometric series formula:

`S_n=(a_1)/(1-r)`

`120=(a_1)/(1-0.75)`

`120=(a_1)/(0.25)`

`120\cdot 0.25=(a_1)/(0.25)\cdot 0.25`

`30=a_1`

Therefore, the value of the first term is a₁ = 30.

Now we have found the first term, we can substitute the values of a₁ and r into the general equation of a geometric sequence.

`\boxed{\begin{minipage}{5.5 cm}\underline{Geometric sequence}\\\\$a_n=a_1\cdot r^(n-1)$\\\\where:\\\phantom{ww}$\bullet$ $a_1$ is the first term. \\\phantom{ww}$\bullet$ $r$ is the common ratio.\\\phantom{ww}$\bullet$ $a_n$ is the $n$th term.\\\phantom{ww}$\bullet$ $n$ is the position of the term.\\\end{minipage}}`

Therefore, the nth term of the geometric sequence is:

`a_n=30 \cdot (0.75)^(n-1)`

To find the value of the third term (a₃), we can substitute n = 3 into the nth term equation:

`a_3=30 \cdot (0.75)^(3-1)`

`a_3=30 \cdot (0.75)^(2)`

`a_3=30 \cdot 0.5625`

`a_3=16.875`

So, the third term of an infinite geometric series with a sum of 120 and a common ratio of 0.75 is 16.875.