Question

find the 18th term of a G.P whose 5th term is 1 and common ratio is2/3​

Answer 1

EXPLANATION.

The G.P whose 5th term is 1.

Common ratio = 2/3.

As we know that,

Nth term of an G.P.

⇒ aₙ = arⁿ⁻¹.

5th term is = ar⁴.

⇒ ar⁴ = 1.

⇒ a(2/3)⁴ = 1.

⇒ a(16/81) = 1.

⇒ 16a/81 = 1.

⇒ 16a = 81.

⇒ a = 81/16.

18th term of the G.P.

⇒ a₁₈ = ar¹⁸⁻¹.

⇒ a₁₈ = ar¹⁷.

⇒ a₁₈ = 81/16(2/3)¹⁷.

⇒ a₁₈ = (3)⁴/(2)⁴(2/3)¹⁷.

⇒ a₁₈ = (2/3)¹³.

                                                                                                                 

MORE INFORMATION.

General term of a G.P.

General term (nth term) of a G.P a + ar + ar²+ ⇒ is given by,

Tₙ = arⁿ⁻¹.

Sum of n terms of a G.P.

The sum of first n terms of an G.P. is given by

Sₙ = a(1 - rⁿ)/1 - r = a - rTₙ/1 - r when r < 1.

Sₙ = a(rⁿ - 1)/r - 1 = rTₙ - a/r - 1. when r > 1.

Sₙ = nr  when  r = 1.

Sum of an infinite G.P.

The sum of an infinite G.P with first term a and common ratio r (-1 < r < 1 that is |r| < 1).

S∞ = a/1 - r.

Geometrical mean (G.M.)

If G is the G.M between two numbers a and b then,

G² = ab ⇒ G = √ab.

Answer 2

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${\large{\bold{\rm{\underline{Given \; that}}}}} \; \; \; \; \red \bigstar$

${\sf{:\implies 5th \: term \: of \: GP \: = 1}}$

${\sf{:\implies Common \: ratio \: = 2/3}}$

${\large{\bold{\rm{\underline{To \; find}}}}} \; \; \; \; \red \bigstar$

${\sf{:\implies 18th \: term \: of \: G.P}}$

${\large{\bold{\rm{\underline{Solution}}}}} \; \; \; \; \red \bigstar$

${\sf{:\implies 18th \: term \: of \: G.P \: is \: (2/3)^{13}}}$

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${\large{\bold{\rm{\underline{Using \; concept}}}}} \; \; \; \; \red \bigstar$

${\sf{:\implies 9th \: term \: of \: GP \: is \: given \: by \: what?}}$

${\large{\bold{\rm{\underline{Using \; formula}}}}} \; \; \; \; \red \bigstar$

${\sf{:\implies a_{n} \: = ar^{n-1}}}$

${\tt{Where,}} \; \; \; \dag$

$\; \; \; \; \; \; \;{\bf{\longmapsto a \: denotes \: first \: term}}$

$\; \; \; \; \; \; \;{\bf{\longmapsto r \: denotes \: common \: ratio}}$

${\tt{Here,}} \; \; \; \dag$

$\; \; \; \; \; \; \;{\bf{\longmapsto ar^{4} \: is \: 5th \: term}}$

$\; \; \; \; \; \; \;{\bf{\longmapsto r \: is \: 2/3}}$

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${\large{\bold{\rm{\underline{Full \; Solution}}}}} \; \; \; \; \red \bigstar$

~ Let us find the first term of GP first..!

${\sf{:\implies ar^{4} \: = 1}}$

${\sf{:\implies a(2/3)^{4} \: = 1}}$

• Law of exponents..!

${\sf{:\implies a(16/81) \: = 1}}$

${\sf{:\implies 16a \: = 1 \times 81}}$

${\sf{:\implies 16a \: = 81}}$

${\sf{:\implies a \: = 81/16}}$

~ Now let's find the 18th term of GP..!

${\sf{:\implies a_{18} \: = ar^{18-1}}}$

${\sf{:\implies a_{18} \: = ar^{17}}}$

${\sf{:\implies a_{18} \: = 81/16(2/3)^{17}}}$

• Law of exponents (Let's break the power's)

${\sf{:\implies a_{18} \: = (3)^{4} / (2)^{4} (2/3)^{17}}}$

• Solving powers..!

${\sf{:\implies a_{18} \: = (2/3)^{13}}}$

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