Question

The value of the first term of a G.P if its sum of three numbers is 7 and their
product is 8.* find the gp.

Answer 1

Answer:-

Given:

Sum of three terms in a GP = 7

Product of the numbers = 8

Let the three numbers be a/r , a , ar where a is the first term and r is the common ratio.

So,

⟹ a/r + a + ar = 7 -- equation (1)

And,

⟹ (a/r) * a * ar = 8

⟹ a³ = 8

⟹ a = ³√8

⟹ a = 2

Substitute the value a in equation (1).

⟹ 2/r + 2 + 2r = 7

⟹ (2 + 2r + 2r²)/r = 7

⟹ 2 + 2r + 2r² = 7r

⟹ 2r² + 2r - 7r + 2 = 0

⟹ 2r² - 5r + 2 = 0

⟹ 2r² - 4r - r + 2 = 0

⟹ 2r(r - 2) - 1 (r - 2) = 0

⟹ (2r - 1)(r - 2) = 0

★ 2r - 1 = 0

⟹ 2r = 1

⟹ r = 1/2

★ r - 2 = 0

⟹ r = 2

If r = 1/2 ;

• a/r = 2/(1/2) = 2 * 2 = 4
• a = 2
• ar = 2 * 1/2 = 1

If r = 2 ;

• a/r = 2/2 = 1
• a = 2
• ar = 2 * 2 = 4

∴ Required GP is 4 , 2 , 1 or 1 , 2 , 4.

Answer 2

${\huge{ \boxed{ \tt{ \underline{question}}}}}$

$\cr \tt \underline{The \: value \: of \: the \: first \: term \: of \: a \: G.P \: if \: its \: sum} \\ \cr \tt\underline{ of \: three \: numbers \: is \: 7 and \: their</p><p>product \: is 8. \: } \\ \cr \tt \underline{ find \: the \: gp.}$

${ \huge{ \boxed{ \tt{ \underline{given}}}}}$

$\\ \tt{Let \: the \: three \: numbers \: be \: \frac{a}{r} \: ,a, \: ar \: } \\ \tt{where \: a \: is \: the \: first}$

$\tt{term \: and \: r \: is \: the \: common \: ratio.} \\ \\ </p><p></p><p> \tt{So,} \\ \\ </p><p></p><p> \tt{= \frac{a}{r} + a + ar = 7 -- equation (1)} \\ \\ </p><p></p><p> \tt{And,} \\ \\ </p><p></p><p> \tt{( \frac{a}{r} ) \times a \times ar = 8} \\ \\ </p><p></p><p> \tt{ {a}^{3} = 8} \\ \\ </p><p></p><p> \tt{= a = \sqrt[3]{8} } \\ \\ </p><p></p><p> \tt{ a = 2}$

$\tt{Substitute \: the \: value \: a \: in \: equation \: (1).}$

$\tt{ \frac{2}{r} + 2 + 2r = 7} \\ \\ </p><p></p><p> \tt{= \frac{(2 + 2r + 2r)}{r} = 7} \\ \\ </p><p></p><p> \tt{2 + 2r + 2r = 7r} \\ \\ </p><p></p><p> \tt{ {2r}^{2} + 2r - 7r + 2 = 0} \\ \\ </p><p></p><p> \tt{- {2r}^{2} - 5r + 2 = 0} \\ \\ </p><p></p><p> \tt{ {2r}^{2} - 4r - r+ 2 = 0} \\ \\ </p><p></p><p> \tt{2r(r- 2)-1 (r- 2) = 0} \\ \\ </p><p></p><p> \tt{(2r - 1)(r - 2) = 0} \\ \\ </p><p></p><p> \tt{2r - 1= 0} \\ \\ </p><p></p><p> \tt{ 2r = 1} \\ \\ </p><p></p><p> \tt{r = \frac{1}{2}} \\ \\ </p><p></p><p> \tt{r- 2 = 0} \\ \\ </p><p></p><p> \tt{r= 2}$

$\tt{ \frac{a}{r} = \frac{2}{ \frac{1}{2} } = 2 \times 2 = 4} \\ \\ </p><p></p><p> \tt{ a = 2} \\ \\ </p><p></p><p> \tt{ar = 2 \times \frac{1}{2} = 1} \\ \\ </p><p></p><p> \tt{If r = 2;} \\ \\ </p><p></p><p> \tt{ \frac{a}{r} = \frac{2}{2} = 1} \\ \\ </p><p></p><p> \tt{a = 2} \\ \\ </p><p></p><p> \tt{ar = 2 \times 2 = 4} \\ \\ </p><p></p><p> \tt{Required \: GP \: is \: 4, 2,1 or 1, 2, 4.}$