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Find the sum of the following geometric series: 35+452+353+454+.... to 2n terms; - Mathematics

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प्रश्न

Find the sum of the following geometric series:

`3/5 + 4/5^2 + 3/5^3 + 4/5^4 + ....` to 2n terms;

योग
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उत्तर

Common Ratio = r = `(3/5)/(4/5^2) = 3/5 × 25/4 = 15/4`

∴ Sum of GP for n terms = `[a(r^n - 1)]/(r - 1)`   ...(1)

⇒ a = `3/5, r = 15/4`, n = 2n

∴ Substituting the above values in (1), we get,

⇒ `[a(r^n - 1)]/(r - 1)`

⇒ `{3/5[(15/4)^"2n" - 1]}/(15/4 - 1)`

⇒ `{3/5[(15/4)^"2n" - 1]}/((15 - 1)/4)`

⇒ `{3/5[(15/4)^"2n" - 1]}/((11)/4)`

⇒ `{3[(15/4)^"2n" - 1]× 4}/(5 × 11)`

⇒ `{12[(15/4)^"2n" - 1]}/(55)`

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अध्याय 20: Geometric Progression - Exercise 20.3 [पृष्ठ २७]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 20 Geometric Progression
Exercise 20.3 | Q 2.5 | पृष्ठ २७

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