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Q 2 - Mathematics

Sum

Q 2

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Solution

Let a be the first term and r be the common ratio og a G.P.
2nd term, t2 = ar = 9
⇒ r = `9/a`

Sum of its infinite terms, S = 48

`=>a/(1-r)=48`

`=>a/(1-9/a)=48`

`=> a^2/(a-9)=48`

⇒ a2 = 48a - 432
⇒ a2 - 48a + 432 = 0
⇒ a2 - 36a - 12a + 432 = 0
⇒ a(a- 36) - 12(a - 36) = 0
⇒ (a- 36)(a - 12) = 0
⇒ a = 36 or a = 12

When a = 36, r = `9/36=1/4`

⇒ 1st term = 36,
2nd term = ar = `36xx1/4=9`

3rd term = ar2 = `36xx1/16=9/4`

When a = 12, r = `9/12=3/4`

⇒ 1st term = 12,

2nd term = ar = `12xx3/4=9`

3rd term = ar2 = `12xx9/16=27/4`

Concept: Geometric Progression - Finding Sum of Their First ‘N’ Terms
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