Question

If the fourth , seventh and tenth terms of a geometrical progression are A , B and C respectively. Then prove that b² = ac.

Answer 1

f = first term

r = common ratio

$a = T_4 = fr^3 \\ b = T_7 = fr^6 \\ c = T_{10} = fr^9 \\ \\ \\ b^2 = (fr^6)^2 \\ \\ = f^2 \times r^{12} \\ \\ = f \times f \times r^{3 + 9} \\ \\ = f \times f \times r^3 \times r^9 \\ \\ = f \times r^3 \times f \times r^9 \\ \\ = (fr^3) \times (fr^9) = ac \\ \\ \\ \therefore b^2 = ac$

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Answer 2

Let the first term of the GP is a and second term is ar and further more terms like this,



Fourth term = A = $ar^{ 4 - 1 } = ar^{3}$


Seventh term = B $= ar^{ 7 - 1 } = ar^{6}$


Tenth term = C = $ar^{ 10 - 1 } = ar^{9}$




To Prove : B² = AC



Proof : Left Hand Side -

= > B²
= > $(ar^{6})^{2}$
= > $a^{2}r^{12}$




Right Hand Side -

= > AC
= > $ar^{3} \times ar^{9}$
= > $a \times r^{3} \times a \times r^{9}$
= > $a^{2}r^{12}$




As Left Hand Side is equal to Right Hand Side.
B² is equal to AC