Question

Find 3 numbers in G.P. such that their sum is 21 and sum of squares is 189.
➥Class 11th.
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➥Answer in detail...no short answers.
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Thanks in advance...!​

Answer 1

Let three numbers in G.P. be a, ar, ar²

$\large\rm { \therefore a + ar + ar^{2} = 21}$

$\large\rm { \implies a( 1 + r + r^{2}) = 21 \ .....(1)}$

Sum of squares = 189

$\large\rm { \therefore a^{2} ( 1 + r^{2} + r^{4} ) = 189 \ .....(2)}$

Squaring both sides of (1) we get

$\large\rm { a^{2} ( 1 + r + r^{2} ) ^{2} = 441 \ .....(3)}$

Dividing (3) by (2) ,

we get $\large\rm { r = \frac{5 \pm 3}{4}}$

$\large\rm { = 2, \frac{1}{2}}$

So, there are two outcomes

when r = 2 from (1)

$\large\rm { a(1+2+4) = 21}$

$\large\rm { a = 3}$

So, numbers are 3,6,12

When r = ½ from (1)

$\large\rm { a = \frac{21 \times 4}{7} }$

$\large\rm { a = 12}$

So, numbers are 12,6,3

Answer 2

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