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Q 6 - ICSE Class 10 - Mathematics

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Question

Q 6

Solution

Let the number be a, ar and ar2.
⇒ (a)2 + (ar)2 + (ar2)2 = 189
⇒ a2 + a2r2 + a2r4 = 189
And, a + ar + ar2 = 21
⇒ (a + ar + ar2) = 212
⇒ a2 + a2r2 + a2r4 + 2a2r + 2a2r3 + 2a2r2 = 441    
⇒ 189 + 2ar(a + ar2 + ar) = 441
⇒ 2ar x 21 = 441 - 189
⇒ 42ar = 252
⇒ ar = 6
⇒ r = `6/a`
Now, a + ar + ar2 = 21

⇒ a + a x `6/a` + a x `36/a^2 = 21`

⇒ a + 6 + `36/a` = 21

⇒ a2 - 6a + 36 = 21a

⇒ a2 - 15a + 36 = 0

⇒ a2 - 12a - 3a + 36 = 0

⇒ a(a - 12) - 3(a - 12) = 0

⇒ (a - 12)(a - 3) = 0

⇒ a = 12 or a = 3

⇒ r = `6/12= 1/2` or r`6/3=2`

Thus, required terms are :

a, ar, ar2 = 12, 12 x `1/2`, 12 x `1/4` OR 3, 3 x 2, 3 x 4

        = 12, 6, 3     OR      3, 6, 12

  Is there an error in this question or solution?
Solution Q 6 Concept: Geometric Progression - Finding Sum of Their First ‘N’ Terms.
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