In an increasing geometric progression of positive terms, the sum of the second and sixth …

Question

In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $$\frac{70}{3}$$ and the product of the third and fifth terms is 49. Then the sum of the $$4^{\text {th }}, 6^{\text {th }}$$

Accepted Answer

Let's denote the first term of the geometric progression by $a$ and the common ratio by $r$. The terms of the geometric progression can be written as follows:

First term: $a$

Second term: $ar$

Third term: $(a{r}^{2})$

Fourth term: $(a{r}^{3})$

Fifth term: $(a{r}^{4})$

Sixth term: $(a{r}^{5})$

Eighth term: $(a{r}^{7})$

We are given two key pieces of information:

1. The sum of the second and sixth terms is $\frac{70}{3}$:

$ar+a{r}^{5}=\frac{70}{3}$

2. The product of the third and fifth terms is 49:

$(a{r}^{2})\cdot (a{r}^{4})=49$

${a}^{2}{r}^{6}=49$

${a}^{2}=\frac{49}{{r}^{6}}$

$a=\frac{7}{{r}^{3}}$

Substituting $a=\frac{7}{{r}^{3}}$ into the first equation:

$\frac{7}{{r}^{3}}\cdot r+\frac{7}{{r}^{3}}\cdot {r}^{5}=\frac{70}{3}$

$\frac{7r}{{r}^{3}}+\frac{7{r}^{5}}{{r}^{3}}=\frac{70}{3}$

$\frac{7}{{r}^{2}}+\frac{7{r}^{2}}{1}=\frac{70}{3}$

Let $x={r}^{2}$. Then:

$\frac{7}{x}+7x=\frac{70}{3}$

Multiply through by 3x to clear the denominator:

$21+21{x}^{2}=70x$

Rearrange into a standard quadratic equation:

$21{x}^{2}-70x+21=0$

Divide by 7 to simplify:

$3{x}^{2}-10x+3=0$

Solve this quadratic equation using the quadratic formula:

$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

Where $a=3$, $b=-10$, and $c=3$. Thus:

$x=\frac{10\pm \sqrt{100-36}}{6}$

$x=\frac{10\pm \sqrt{64}}{6}$

$x=\frac{10\pm 8}{6}$

$x=3$ or $x=\frac{1}{3}$

Since $x={r}^{2}$ and $r$ is positive, we get $r=\sqrt{3}$ or $r=\frac{1}{\sqrt{3}}$. We need to choose the value that results in positive, increasing terms:

If $r=\sqrt{3}$:

$a=\frac{7}{{r}^{3}}=\frac{7}{(\sqrt{3}{)}^{3}}=\frac{7}{3\sqrt{3}}=\frac{7}{3}\cdot \frac{1}{\sqrt{3}}=\frac{7\sqrt{3}}{9}$

Now we can determine the sum of the 4th, 6th, and 8th terms:

The 4th term is: $a{r}^{3}=\frac{7\sqrt{3}}{9}\cdot 3\sqrt{3}=7$

The 6th term is: $a{r}^{5}=\frac{7\sqrt{3}}{9}\cdot 9\sqrt{3}=21$

The 8th term is: $a{r}^{7}=\frac{7\sqrt{3}}{9}\cdot 27(\sqrt{3})=49$

Adding these together:

$(4th+6th+8thterms)=7+21+63=91$

Therefore, the sum of the 4th, 6th, and 8th terms is 91.

Correct answer:

Option C: 91

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