Sum of an infinite GP MCQ Quiz in বাংলা - Objective Question with Answer for Sum of an infinite GP - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

If a, ar, ar2, …. is an infinite GP, then the sum of infinite geometric series is given by.

Sum of infinite GP = a/1 - r ; |r| < 1

Calculation:

S = \(\frac{1}{7}+\frac{2}{7^{2}}+\frac{1}{7^{3}}+\frac{2}{7^{4}}\) + ....

= (\(\frac{1}{7}+\frac{1}{7^{3}}+\frac{1}{7^{5}}\)+.....) + (\(\frac{2}{7^{2}}+\frac{2}{7^{4}}+\frac{2}{7^{6}}+...\)) 

= S1 + S2 

S1 = (\(\frac{1}{7}+\frac{1}{7^{3}}+\frac{1}{7^{5}}\) +.....) 

In the first series a = 1/7  and r =  1/49

Now; S1  = a/1 - r =  7/48

S2 = (\(\frac{2}{7^{2}}+\frac{2}{7^{4}}+\frac{2}{7^{6}}+...\)) 

In the second series a = 2/49 and r = 1/49

Now, S2 = a/1 - r = 2/48

S = S1 + S2  = 9/48 = 3/16

Concept:

The sum of an infinite G.P. is 

S = \(\rm a\over 1-r\)

Where a is the first term of the series and r is the common ratio

Calculation:

Given 1 + sin θ + sin2 θ + ... upto ∞ = 2√3 + 4

First term of the G.P is a = 1 and r = sin θ 

∴ \(\rm 1\over 1 - \sin θ\) = 2√3 + 4

1 - sin θ = \(1\over2\sqrt3 + 4\)

1 - sin θ = \({1\over2\sqrt3 + 4} \times {2\sqrt3 - 4\over 2\sqrt3 - 4}\)

1 - sin θ = \( {4-2\sqrt3 \over 16-12}\)

1 - sin θ = \( 1-{\sqrt3 \over 2}\)

sin θ = \( {\sqrt3 \over 2}\) 

⇒ θ = \(\boldsymbol{\pi\over3}\)

Given :

x = 1 + a + a2 + a3 + ...∞ (|a| < 1)

y = 1 + b + b2 + b3 + ...∞ (|b| < 1)

Concept :

Sum of the infinite term of GP:

S = \(\frac{a}{1-r}\)     

Where a is first term and r is common ratio and |r| < 1

Calculations :

x = 1 + a + a2 + a3 + ...∞ (|a| < 1)

First-term = 1 and common ratio = a

By using the above concept

⇒ \(x=\frac{1}{1-a}\)

⇒ x - ax  = 1

⇒ \(a = \frac{x -1}{x}\)        ----- (1)

Similarly, we get

⇒ \(b = \frac{y-1}{y}\)        ----- (2)

Now, consider the series  1 + ab + a2b2 + a3b3 + ...∞

Using equation (1), we get

⇒ \(z= \frac{1}{1-ab}\)

Put the value of a and b from equations (1) and (2), we get

⇒ \(z= \frac{1}{1-\frac{x-1}{x}.\frac{y-1}{y}}\)

⇒ \(z = \frac{xy}{x+y-1}\)

∴ The required value is equal to \(\frac{xy}{x \ + \ y \ - \ 1}\).

Concept:

Let us consider sequence a₁, a₂, a₃ …. a_n is an G.P.

• Common ratio = r = \(\frac{{{{\rm{a}}_2}}}{{{{\rm{a}}_1}}} = \frac{{{{\rm{a}}_3}}}{{{{\rm{a}}_2}}} = \ldots = \frac{{{{\rm{a}}_{\rm{n}}}}}{{{{\rm{a}}_{{\rm{n}} - 1}}}}\)
• n^th  term of the G.P. is an = ar^n-1
• Sum of n terms = s = \(\frac{{{\rm{a\;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{\;}}1}}\); where r >1
• Sum of n terms = s = \(\frac{{{\rm{a\;}}\left( {1 - {\rm{\;}}{{\rm{r}}^{\rm{n}}}} \right)}}{{1 - {\rm{\;r}}}}\); where r <1
• Sum of infinite GP = \({{\rm{s}}_\infty } = \frac{{\rm{a}}}{{1{\rm{\;}} - {\rm{\;r}}}}\); |r| < 1

Calculation:

\({\rm{Let\;S}} = {6^{\frac{1}{2}}} \times {6^{\frac{1}{2}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{1}{4}}} \times \ldots \)

This can be written as:

\({\rm{S}} = {6^{\frac{1}{2}}} \times {6^{\frac{2}{4}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{4}{{16}}}} \times \ldots \)

\(S = {6^{\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{{16}} + \ldots }}\)

Let \(x = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{{16}} + \ldots\)            ---(1)

Multiply by 1/2 on both sides, we get:

\(\Rightarrow \frac{x}{2} = \frac{1}{4} + \frac{2}{8} + \frac{3}{{16}} + \frac{4}{{32}} + \ldots \)            ---(2)

Subtract equation 2 from equation 1, we get

\(x - \frac{x}{2} = \left( {\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{{16}} + \ldots } \right) - \;\left( {\frac{1}{4} + \frac{2}{8} + \frac{3}{{16}} + \frac{4}{{32}} + \ldots } \right)\)

\(\Rightarrow \frac{x}{2} = \frac{1}{2} + \left( {\frac{2}{4} - \frac{1}{4}} \right) + \left( {\frac{3}{8} - \frac{2}{8}} \right) + \left( {\frac{4}{{16}} - \frac{3}{{16}}} \right) + \ldots \)

\(\Rightarrow \frac{x}{2} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \ldots \)

The above series is clearly a Geometric Progression with the first term = 1/2 and common ratio = 1/2

\(\therefore \frac{x}{2} = \frac{{\frac{1}{2}}}{{\left( {1 - \frac{1}{2}} \right)}} = \frac{{\frac{1}{2}}}{{\frac{1}{2}}} = 1\)

⇒ x = 2 × 1 = 2

Put the value of x in the original sum:

∴ S = 6² = 36

Concept:

If a, ar, ar², ….is terms an infinite GP, then the sum of infinite geometric series is given by.

Sum of infinite GP = \({s_\infty } = \;\frac{a}{{1\; - \;r}}\) ; |r| < 1

Calculation:

If a, ar, ar², ….is terms an infinite GP.

Given: Second term of GP = 2

⇒ ar = 2      …. (1)

Sum of infinite GP = \({s_\infty } = \;\frac{a}{{1\; - \;r}} = 8\)

⇒ a = 8 × (1 – r)

Multiplying r both sides,

⇒ ar = 8r × (1- r)

From equation 1^st

⇒ 2 = 8 (r – r²)

⇒ 1 = 4r – 4r²

⇒ 4r² – 4r + 1 = 0

⇒ 4r² – 2r – 2r + 1 = 0

⇒ 2r (2r – 1) – 1(2r – 1) = 0

⇒ (2r – 1) (2r – 1) = 0

⇒ 2r – 1 =0

So, r = ½

Put in equation 1^st

a × ½ = 2

So, a = 4

Then the GP is a, ar ,ar²…….

So 4,2,1…..are in GP

Concept Used:

a, ar, ar², ar³, . . ., ar^n-1 is a G.P. where a is the first term and r is the common ratio.

and for any quadratic ax² + bx + c = 0, 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Calculation:

Given: t_n = t_n+1 + t_n+2

⇒ ar^n-1 =  arⁿ + arⁿ⁺¹

⇒ arn-1 = ar^n-1( r+ r²)

⇒ 1 = r+ r2

⇒ r2 + r -1 = 0

⇒ \(r = \frac{-1± \sqrt{1^2- 4(-1)(1)}}{2}\)

⇒ \(r = \frac{-1± \sqrt{1+ 4}}{2}\)

⇒ \(r = \frac{-1± \sqrt{5}}{2}\)

Hence, option (4) is the the correct answer.

Concept:

A square is drawn by joining mid-points of the sides of a square. Another square is drawn inside the second square in the same way and the process is continued indefinitely. 

The side of the first square = a 

Side of the second square = \(\rm \dfrac a {\sqrt2}\) and so on....

If the series is in GP with common ratio r and first term a then sum of infinity GP = \(\rm \dfrac a {1-r}\)

Calculations:

Given, a square is drawn by joining mid-points of the sides of a square. Another square is drawn inside the second square in the same way and the process is continued infinity.

The side of the first square is 16 cm.

Side of second square = \(\rm \dfrac {16}{\sqrt2}\) = 8√2 cm.

Side of third square = 8 cm. and so on... 

Sum of area of squares = (16)² + (8 √2)² + (8)² + ....

Sum of area of squares = 256 + 128 + 64 + ....

The series is in GP

⇒Sum of the area of squares = \(\dfrac {256}{1-\dfrac12}\)

⇒Sum of the area of squares = 512 

Concept Used:

Sum of an infinite geometric series: \(\frac{a}{1-r}\), where |r| < 1

Second term of a geometric series: ar

Calculation

Given, \(\frac{a}{1-r} = 4\)...(1)

Also given, \(ar = \frac{3}{4}\)...(2)

From (1), \(a = 4(1-r)\)

Substituting in (2),

\(4(1-r)r = \frac{3}{4}\)

⇒ \(4r - 4r^2 = \frac{3}{4}\)

⇒ \(16r - 16r^2 = 3\)

⇒ \(16r^2 - 16r + 3 = 0\)

⇒ \((4r - 3)(4r - 1) = 0\)

⇒ \(r = \frac{3}{4}\) or \(r = \frac{1}{4}\)

If \(r = \frac{3}{4}\), then from (2), \(a(\frac{3}{4}) = \frac{3}{4} \implies a = 1\)

If \(r = \frac{1}{4}\), then from (2), \(a(\frac{1}{4}) = \frac{3}{4} \implies a = 3\)

∴ \(a = 1, r = \frac{3}{4}\) or \(a = 3, r = \frac{1}{4}\)

Hence option 2 are correct

Concept:

In a random experiment, let S be the sample space and let E ⊆ S. Then, E is an event.

The probability of occurrence of E is defined as,

\(\rm P(E)=\frac{n(E)}{n(S)}\)where, n(E) = number of elements in E and n(S) = number of possible outcomes.

Sum of infinite term in G.P. = \(\rm \frac {a} {(1 - r)}\)

Here a = first term, r = common ratio

Calculation:

n(s) = {1, 2, 3, 4, 5, 6} = 6

Odd number = {1, 3, 5} = 3

Probability of odd number = \(\rm \frac {3} {6} = \frac {1} {2}\)

Winning probability = \(\rm \frac{1}{2}\)

Lossing probabily = \(\rm \frac{1}{2}\)

If B play first so conditions are:

(B loss)(A win) + (B loss)(A loss)(B loss)(A win) + (B loss)(A loss)(B loss)(A loss)(B loss)(A win) + ...+ infinite

= \(\rm [{\frac {1}{2} \times \frac {1}{2}}] + [{\frac {1}{2} \times \frac {1}{2} \times \frac {1}{2} \times \frac {1}{2}}] + [{\frac {1}{2} \times \frac {1}{2} \times \frac {1}{2} \times \frac {1}{2}\times \frac {1}{2} \times \frac {1}{2}}] +\cdots + \infty \)

Take common \(\rm 1\over4\)

= \(\rm \frac{1} {4} [1 +{\frac {1}{2} \times \frac {1}{2}} + {\frac {1}{2} \times \frac {1}{2} \times \frac {1}{2} \times \frac {1}{2}} + {\frac {1}{2} \times \frac {1}{2} \times \frac {1}{2} \times \frac {1}{2}\times \frac {1}{2} \times \frac {1}{2}}+\cdots+ \infty] \)

Here a = 1, r = \(\rm 1\over4\)

= \(\rm \frac {1} {4} \times \frac {1} {(1 - \frac {1} {4})}\)

= \(\rm \frac {1} {4} \times \frac {(4 \;\times\; 1)} {3}\)

= \(\rm 1\over 3\)

Calculation

\(\sum_{\mathrm{n}=0}^{∞} \mathrm{ar}^{\mathrm{n}}\) = 57

⇒ a + ar + ar² + ∞ = 57

⇒ \(\frac{\mathrm{a}}{1-\mathrm{r}}=57 \rm \quad ...(I)\)

\(\rm \sum_{n=0}^{∞} a^3 r^{3 n}\) = 9747

⇒ a³ + a³. r³ + a³. r⁶ + .... ∞ = 9746

⇒ \(\frac{a^3}{1-r^3}=9746 \rm \quad ...(II)\)

From (I) and (II)

\(\frac{(\mathrm{I})^3}{\text { (II) }} \Rightarrow \frac{\frac{\mathrm{a}^3}{(1-\mathrm{r})^3}}{\frac{\mathrm{a}^3}{1-\mathrm{r}^3}}=\frac{57^3}{9717}=19\)

On solving, r = \(\frac{2}{3}\) and r = \(\frac{3}{2}\) (rejected)

⇒ a = 19

∴ a + 18r = 19 + 18 × \(\frac{2}{3}\) = 31

Hence option (4) is correct