Question

Which term of the sequence 1/2, -1/4, 1/8,-1/16,.... is -1/256 ?​

Answer 1

EXPLANATION.

Sequence = 1/2,-1/4,1/8,-1/16+.-1/256.

As we know that,

First term of a GP = a = 1/2.

Common Ratio of GP = b/a.

⇒ -1/4/1/2 = -1/4 X 2/1 = -1/2.

⇒ 1/8/-1/4 = 1/8 X 4/-1 = -1/2.

General term of a G.P

⇒ a + a(r) + ar² + ar³ + = Tₙ = a(rⁿ⁻¹).

⇒ Tₙ term = -1/256.

Using the formula, we get.

⇒ Tₙ = a(rⁿ⁻¹).

⇒ -1/256 = (1/2).(-1/2)ⁿ⁻¹.

⇒ (-1/2⁸) = (-1/4)ⁿ⁻¹.

⇒ 2⁸ = 2²⁽ⁿ⁻¹⁾.

⇒ 8 = 2n - 2.

⇒ 10 = 2n.

⇒ n = 5.

Their term of the G.P = 5th term.

                                                                                         

MORE INFORMATION.

(1) = General term of a G.P.

General term (nth term) of a G.P a + a(r) + ar² +  = Tₙ = a(rⁿ⁻¹).

(2) = 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 - (r)(Tₙ)/1 - r where r < 1.

Sₙ = a(rⁿ - 1)/r - 1 = r(Tₙ) - (a)/r - 1. where r > 1.

Sₙ = n r where r = 1.

(3) = Sum of an infinite G.P.

The Sum of an infinite G.P with first term a and common ratio r,

r(-1 < r < 1  Or | r | < 1 ) is S∞ = a/1 - r.

Answer 2

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

★ Sequence of G.P is given as ➝ 1/2, -1/4, 1/8, -1/16 .... = -1/256

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

★ 5th them of G.P

${\large{\bold{\rm{\underline{Solution}}}}}$

★ 5th them of G.P = 5

${\large{\bold{\rm{\underline{Using \; concepts}}}}}$

★ The common ratio of G.P is given by ?

★ The general term of GP is given by ?

${\large{\bold{\rm{\underline{Using \; dimensions}}}}}$

★ The common ratio of G.P is given by ➝ b/a

★ The general term of GP is given by ➝ ${\sf{a + a(r) + ar^{2} + ar^{3} = T_{n} = a(r^{n-1})}}$

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

__________________

~ According to the question we can see that the 1st term of the sequence is 1/2

Henceforth,

${\frak{The \: common \: ratio \: (G.P) \: is \: b/a}}$

~ According to the question and using the formula of the common ratio of G.P let's put the values,

✠ -1/4/1/2

• Let us cross multiply !..

✠ -1/4 × 2/1

• Multiplying !.

✠ -2/4

• Cancelling the digits !..

✠ -1/2

__________________

~ Now let's see what to do !..

✠ 1/8/-1/4

• Let us cross multiply !..

✠ 1/8 × 4/-1

• Let's multiply !..

✠ 4/8

• Let us cancel !..

✠ 1/2

__________________

${\frak{General \: term \: of \: GP \: is \: given \: by \: \rightarrow a + a(r) + ar^{2} + ar^{3} = T_{n} = a(r^{n-1})}}$

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

${\sf{\mapsto T_{n} \: term \: is \: -1/256}}$

~ Let's put the values, according to the main formula..!

✠ ${\sf{a(r^{n-1})}}$

✠ ${\sf{-1/256 \: = \: (1/2)(-1/2)^{n-1}}}$

✠ ${\sf{-1/256 \: = \: (-1/4)^{n-1}}}$

• Law of exponents !..

✠ ${\sf{(-1/2^{8}) \: = \: (-1/4)^{n-1}}}$

• Again law of exponents !..

• (+ = -) ; (- = +)

✠ ${\sf{2^{8} \: = \: 2^{2(n-1)}}}$

• Again law of exponents !.. and cancelling the digits !..

✠ ${\sf{8 \: = \: 2n \: - 2}}$

• (+ = -) ; (- = +)

✠ ${\sf{8 + 2 \: = 2n}}$

✠ ${\sf{10 \: = 2n}}$

✠ ${\sf{10/2 = n}}$

✠ ${\sf{5 = n}}$

✠ ${\sf{n = 5}}$

${\frak{\red{Henceforth, \: 5 \: in \: term \: of \: GP}}}$

${\large{\bold{\rm{\underline{Additional \; information}}}}}$

Law of Exponents -

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto a^{m} \times a^{n} = \: a^{m+n}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (a^{m})^{n} \: = a^{mn}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto a^{m} \times b^{m} = (ab)^{m}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto a^{0} = 1}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto \dfrac{a^{m}}{b^{m}} = (\dfrac{a}{b})^{m}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto \dfrac{a^{m}}{a^{n}} = a^{m-n}}}}$

Where, m - n ∈ N

Easy to remember about last rule ⬆️

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto x^{a} ÷ x^{b} = x^{a-b}}}}$

It happens when a > b

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto x^{a} ÷ x^{b} = \dfrac{1}{x^{b-a}}}}}$

It's happen when a < b

Law of Exponents -

$\begin{gathered}\begin{gathered}\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\tt\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\tt{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\tt(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\tt\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\tt\sqrt[\tt n]{\tt a} = (a)^{\dfrac{1}{n}}\end{minipage}}\end{gathered}\end{gathered}$