Question

est
Find three numbers in G.P. such that their
sum is 21 and sum of their squares is 189.​

Answer 1

Answer:

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Answer 2

$\large\underline{\sf{Solution-}}$

$\sf \: Let \: three \: numbers \: in \: GP \: be \: a, \: ar, \: a {r}^{2}$

According to statement,

• The sum of three numbers in GP is 21.

$\bf :\longmapsto\:a + ar + {ar}^{2} = 21- - - (1)$

Again,

According to statement,

• The sum of the squares of terms of GP is 189

$\bf :\longmapsto\: {a}^{2} + {a}^{2} {r}^{2} + {a}^{2} {r}^{4} = 189 - - (2)$

Now, Squaring both sides of equation (1), we get

$\rm :\longmapsto\: {(a + ar + {ar}^{2} )}^{2} = {(21)}^{2}$

$\rm :\longmapsto\: {a}^{2} + {a}^{2} {r}^{2} + {a}^{2} {r}^{4} + 2 {a}^{2}r + 2 {a}^{2} {r}^{3} + 2 {a}^{2} {r}^{2} = 441$

$\rm :\longmapsto\:189 + 2ar(a + {ar}^{2} + ar) = 441$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \boxed{ \bf \because \{ \: using \: (2) \}}$

$\rm :\longmapsto\:2ar \times 21 = 441 - 189$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \boxed{ \bf \because \{ \: using \: (1) \}}$

$\rm :\longmapsto\:42ar = 252$

$\rm :\implies\:ar = 6$

$\bf :\implies\:a = \dfrac{6}{r} - - (3)$

Now, from equation (1), we have

$\rm :\longmapsto\:a + ar + {ar}^{2} = 21$

$\rm :\longmapsto\:a(1 + r + {r}^{2} ) = 21$

$\rm :\longmapsto\:\dfrac{6}{r} (1 + r + {r}^{2} ) = 21$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \boxed{ \bf \because \{ \: using \: (3) \}}$

$\rm :\longmapsto\:6 + 6r + {6r}^{2} = 21r$

$\rm :\longmapsto\: {6r}^{2} - 15r + 6 = 0$

$\rm :\longmapsto\: {2r}^{2} - 5r + 2 = 0$

$\rm :\longmapsto\: {2r}^{2} - 4r - r + 2 = 0$

$\rm :\longmapsto\:2r(r - 2) - 1(r - 2) = 0$

$\rm :\longmapsto\:(2r - 1)(r - 2) = 0$

$\bf\implies \:r = \dfrac{1}{2} \: \: \: or \: \: \: r = 2$

So, we get values of 'a' from equation (3) as

$\begin{gathered}\boxed{\begin{array}{c|c} \bf r & \bf a \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 2 & \sf 3 \\ \\ \sf \dfrac{1}{2} & \sf 12 \end{array}} \\ \end{gathered}$

Two cases arises,

Case :-1

$\rm :\longmapsto\:When \: a \: = \: 3 \: and \: r \: = \: 2$

$\begin{gathered}\begin{gathered}\bf \: Numbers \: are - \begin{cases} &\sf{a = 3} \\ &\sf{ar = 2 \times 3 = 6} \\ &\sf{ {ar}^{2} = 3 \times {2}^{2} = 12 } \end{cases}\end{gathered}\end{gathered}$

Case :- 2

$\rm :\longmapsto\:When \: a = 12 \: \: and \: \: r \: = \: \dfrac{1}{2}$

$\begin{gathered}\begin{gathered}\bf \: Numbers \: are - \begin{cases} &\sf{a = 12} \\ &\sf{ar = 12 \times \dfrac{1}{2} = 6} \\ &\sf{ {ar}^{2} = 12 \times {\bigg( \dfrac{1}{2}\bigg) }^{2} = 3} \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

↝ nᵗʰ term of an Geometric sequence is,

$\rm :\longmapsto\:\boxed{ \bf \: a_n = {ar}^{n - 1} }$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.

↝ Sum of n term of an Geometric sequence is,

$\boxed{ \bf \: S_n = \dfrac{a( {r}^{n} - 1)}{r - 1} \: provided \: r \ne \: 1}$

$\boxed{ \bf \: S_n \: = \: n \times a \: \: \: \: provided \: that \: r \: = \: 1}$

$\boxed{ \bf \: S_ \infty = \dfrac{a}{1 - r} \: provided \: that \: |r| < 1}$