LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

GEOMETRIC PROGRESSION - GENERAL TERM OF G.P.

**General Term of G.P.-
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__Theorem.1-__ with 1st term ‘a’, common ratio ‘r’, show that its e-th term is given by Tₑ = arᵉˉ¹

__Proof –__ Let us consider a G.P. with 1st term = a, and common ratio = r.

**Then, the G.P. is a, ar, ar², ar³, ar⁴, ar⁵,………………….
**

**In this G.P. we have
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**1st term T₁ = a = ar⁰ = ar⁽¹ˉ¹⁾
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**2nd term T₂ = ar = ar¹ = ar⁽²ˉ¹⁾
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**3rd term T₃ = ar² = ar⁽³ˉ¹⁾
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**4th term T₄ = ar³ = ar⁽⁴ˉ¹⁾
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** ……………. ……………. ………………
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** ……………. ……………. ………………
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**So, e-th term, Tₑ = ar⁽ᵉˉ¹⁾
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**Hence, its e-th term is given by Tₑ = ar⁽ᵉˉ¹⁾
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**Example.1) Show that the progression 3, 6, 12, 24, 48,…………… is a G.P. write its
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**(i) first term (ii) common ratio
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**(iii) e-th term (iv) 12th term
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**Ans.) The given terms are 3, 6, 12, 24, 48,……………
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** 6 12 24 48
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**Clearly, we have ------ = ------ = ------- = ------- = 2 (constant)
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** 3 6 12 24
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**So, the given number form a G.P.
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**Clearly, we have
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**(i) first term a = 3 (Ans.)
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** 6
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**(ii) common ratio, r = ------ = 2 (Ans.)
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** 3
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** (iii) e-th term, Tₑ = arᵉˉ¹
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** => Tₑ = (3 X 2ᵉˉ¹) ……………….(Ans.)
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**(iv) Putting e = 12, in Tₑ = arᵉˉ¹ where a = 3, r = 2 and we get –
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** Tₑ = arᵉˉ¹
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**=> T₁₂ = (3 X 2¹²ˉ¹)
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** => (3 X 2¹¹)
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** => (3 X 2048)
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** => 6144 ……………(Ans.)**

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