Find four numbers in g.p. such that sum ofthe middle two numbers is 10/3 and theirproduct is 1.

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Find three numbers in G.P. whose sum is 65 and whose product is 3375 .

Find four numbers in $$G.P.$$ such that their products is $$1$$ and the sum of the middle two numbers is $$\cfrac { 10 }{ 3 } $$. Let the terms be $$a,a{r},a{r^{2}},a{r^{3}}$$

Product is $$1\implies a(a{r})(a{r^{2}})(a{r}^{3})=1$$

                   $$\implies a^{4}{r^{6}}=1\implies a=r^{-\dfrac{3}{2}}$$

Sum of middle two terms is $$\dfrac{10}{3}\implies a{r}+a{r^{2}}=\dfrac{10}{3}$$

                                                     $$\implies a{r}(1+r)=\dfrac{10}{3}\implies 1+r=\dfrac{10}{3}\sqrt{r}$$

On solving ,we get $$r=9,\dfrac{1}{9}\implies a=\dfrac{1}{27},27$$

the four numbers in $$G.P.$$ are $$\dfrac{1}{27},\dfrac{1}{3},3,27$$

Let the four numbers in G.P. be `"a"/"r"^3, "a"/"r"`, ar, ar3.

According to the second condition,

`"a"/"r"^3 ("a"/"r") ("ar")("ar"^3)` = 1

∴ a4 = 1∴ a = 1

According to the first condition,

`"a"/"r" + "ar"  10/3`

∴ `1/"r" + (1)"r" = 10/3`

∴ `(1 + "r"^2)/"r" = 10/3`

∴ 3 + 3r2 = 10r
∴ 3r2 – 10r + 3 = 0
∴ (r – 3)(3r – 1) = 0

∴ r = 3 or r = `1/3`

When r = 3, a = 1

`"a"/"r"^3 = 1/(3)^3 = 1/27, "a"/"r" = 1/3, "ar"` = 1(3) = 3 and ar3 = 1(3)3 = 27

When r = `1/3`, a = 1

`"a"/"r"^3 = 1/((1/3)^3) = 27, "a"/"r" = 1/((1/3)) = 3`,

`"ar" = 1(1/3) = 1/3 and "ar"^3 = 1(1/3)^3 = 1/27`

∴ the four numbers in G.P.are

`1/27, 1/3, 3, 27 or 27, 3, 1/3, 1/27`.