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# RD Chapter 20- Geometric Progressions Ex-20.3 Interview Questions Answers

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Question 1 :

Find the sum of the following geometric progressions:

(i) 2, 6, 18, … to 7 terms

(ii) 1, 3, 9, 27, … to 8 terms

(iii) 1, -1/2, ¼, -1/8, …

(iv) (a2 – b2), (a – b), (a-b)/(a+b), … to nterms

(v) 4, 2, 1, ½ … to 10 terms

(i) 2, 6, 18, … to 7 terms

We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)

Given:

a = 2, r = t2/t1 =6/2 = 3, n = 7

Now let us substitutethe values in

a(rn –1)/(r – 1) = 2 (37 – 1)/(3-1)

= 2 (37 –1)/2

= 37 –1

= 2187 – 1

= 2186

(ii) 1, 3, 9, 27, … to 8terms

We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)

Given:

a = 1, r = t2/t1 =3/1 = 3, n = 8

Now let us substitutethe values in

a(rn –1)/(r – 1) = 1 (38 – 1)/(3-1)

= (38 –1)/2

= (6561 – 1)/2

= 6560/2

= 3280

(iii) 1, -1/2, ¼, -1/8, …

We know that, sum ofGP for infinity = a/(1 – r)

Given:

a = 1, r = t2/t1 =(-1/2)/1 = -1/2

Now let us substitutethe values in

a/(1 – r) = 1/(1 –(-1/2))

= 1/(1 + 1/2)

= 1/((2+1)/2)

= 1/(3/2)

= 2/3

(iv) (a2 –b2), (a – b), (a-b)/(a+b), … to n terms

We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)

Given:

a = (a2 –b2), r = t2/t1 = (a-b)/(a2 –b2) = (a-b)/(a-b) (a+b) = 1/(a+b), n = n

Now let us substitutethe values in

a(rn –1)/(r – 1) =

(v) 4, 2, 1, ½ … to 10terms

We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)

Given:

a = 4, r = t2/t1 =2/4 = 1/2, n = 10

Now let us substitutethe values in

a(rn –1)/(r – 1) = 4 ((1/2)10 – 1)/((1/2)-1)

= 4 ((1/2)10 –1)/((1-2)/2)

= 4 ((1/2)10 –1)/(-1/2)

= 4 ((1/2)10 –1) × -2/1

= -8 [1/1024 -1]

= -8 [1 – 1024]/1024

= -8 [-1023]/1024

= 1023/128

Question 2 :

Find the sum of the following geometric series :
(i) 0.15 + 0.015 + 0.0015 + … to 8 terms;

(ii) √2 + 1/√2 + 1/2√2 + …. to 8 terms;

(iii) 2/9 – 1/3 + ½ – ¾ + … to 5 terms;

(iv) (x + y) + (x2 + xy + y2) + (x3 +x2 y + xy2 + y3) + …. to n terms ;

(v) 3/5 + 4/52 + 3/53 + 4/54 +… to 2n terms;

(i) 0.15 + 0.015 + 0.0015+ … to 8 terms

Given:

a = 0.15

r = t2/t1 =0.015/0.15 = 0.1 = 1/10

n = 8

By using the formula,

Sum of GP for n terms= a(1 – rn )/(1 – r)

a(1 – rn )/(1– r) = 0.15 (1 – (1/10)8) / (1 – (1/10))

= 0.15 (1 – 1/108)/ (1/10)

= 1/6 (1 – 1/108)

(ii) √2 + 1/√2 + 1/2√2 + ….to 8 terms;

Given:

a = √2

r = t2/t1 =(1/√2)/√2 = 1/2

n = 8

By using the formula,

Sum of GP for n terms= a(1 – rn )/(1 – r)

a(1 – rn )/(1– r) = √2 (1 – (1/2)8) / (1 – (1/2))

= √2 (1 – 1/256) /(1/2)

= √2 ((256 – 1)/256) ×2

= √2 (255×2)/256

= (255√2)/128

(iii) 2/9 – 1/3 + ½ – ¾ + …to 5 terms;

Given:

a = 2/9

r = t2/t1 =(-1/3) / (2/9) = -3/2

n = 5

By using the formula,

Sum of GP for n terms= a(1 – rn )/(1 – r)

a(1 – rn )/(1– r) = (2/9) (1 – (-3/2)5) / (1 – (-3/2))

= (2/9) (1 + (3/2)5)/ (1 + 3/2)

= (2/9) (1 + (3/2)5)/ (5/2)

= (2/9) (1 + 243/32) /(5/2)

= (2/9) ((32+243)/32)/ (5/2)

= (2/9) (275/32) × 2/5

= 55/72

(iv) (x + y) + (x2 +xy + y2) + (x3 + x2 y + xy2 +y3) + …. to n terms;

Let Sn =(x + y) + (x2 + xy + y2) + (x3 + x2 y+ xy2 + y3) + …. to n terms

Let us multiply anddivide by (x – y) we get,

Sn =1/(x – y) [(x + y) (x – y) + (x2 + xy + y2) (x – y)… upto n terms]

(x – y) Sn =(x2 – y2) + x3 + x2y + xy2 –x2y – xy2 – y3..upto n terms

(x – y) Sn = (x2 +x3 + x4+…n terms) – (y2 + y3 +y4 +…n terms)

By using the formula,

Sum of GP for n terms= a(1 – rn )/(1 – r)

We have two G.Ps inabove sum, so,

(x – y) Sn =x2 [(xn – 1)/ (x – 1)] – y2 [(yn –1)/ (y – 1)]

Sn =1/(x-y) {x2 [(xn – 1)/ (x – 1)] – y2 [(yn –1)/ (y – 1)]}

(v) 3/5 + 4/52 +3/53 + 4/54 + … to 2n terms;

The series can bewritten as:

3 (1/5 + 1/53 +1/55+ … to n terms) + 4 (1/52 + 1/54 +1/56 + … to n terms)

Firstly let usconsider 3 (1/5 + 1/53 + 1/55+ … to n terms)

So, a = 1/5

r = t2/t1 =1/52 = 1/25

By using the formula,

Sum of GP for n terms= a(1 – rn )/(1 – r)

Now, Let us consider 4(1/52 + 1/54 + 1/56 + … to nterms)

So, a = 1/25

r = t2/t1 =1/52 = 1/25

By using the formula,

Sum of GP for n terms= a(1 – rn )/(1 – r)

Question 3 : Evaluate the following:

= (2 + 31)+ (2 + 32) + (2 + 33) + … + (2 + 311)

= 2×11 + 31 +32 + 33 + … + 311

= 22 + 3(311 –1)/(3 – 1) [by using the formula, a(1 – rn )/(1 – r)]

= 22 + 3(311 –1)/2

= [44 + 3(177147 –1)]/2

= [44 + 3(177146)]/2

= 265741

= (2 + 30)+ (22 + 3) + (23 + 32) + … + (2n +3n-1)

= (2 + 22 +23 + … + 2n) + (30 + 31 +32 + …. + 3n-1)

Firstly let usconsider,

(2 + 22 +23 + … + 2n)

Where, a = 2, r = 22/2= 4/2 = 2, n = n

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

= 2 (2n –1)/(2 – 1)

= 2 (2n –1)

Now, let us consider

(30 +31 + 32 + …. + 3n)

Where, a = 30 =1, r = 3/1 = 3, n = n

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

= 1 (3n –1)/ (3 – 1)

= (3n –1)/2

So,

= (2 + 22 +23 + … + 2n) + (30 + 31 +32 + …. + 3n)

= 2 (2n –1) + (3n – 1)/2

= ½ [2n+2 +3n – 4 – 1]

= ½ [2n+2 +3n – 5]

= 42 +43 + 44 + … + 410

Where, a = 42 =16, r = 43/42 = 4, n = 9

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

= 16 (49 –1)/(4 – 1)

= 16 (49 –1)/3

= 16/3 [49 –1]

Question 4 :

Find the sum of the following series :
(i) 5 + 55 + 555 + … to n terms.

(ii) 7 + 77 + 777 + … to n terms.

(iii) 9 + 99 + 999 + … to n terms.

(iv) 0.5 + 0.55 + 0.555 + …. to n terms

(v) 0.6 + 0.66 + 0.666 + …. to n terms.

(i) 5 + 55 + 555 + … to nterms.

Let us take 5 as acommon term so we get,

5 [1 + 11 + 111 + … nterms]

Now multiply anddivide by 9 we get,

5/9 [9 + 99 + 999 + …n terms]

5/9 [(10 – 1) + (102 –1) + (103 – 1) + … n terms]

5/9 [(10 + 102 +103 + … n terms) – n]

So the G.P is

5/9 [(10 + 102 +103 + … n terms) – n]

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

Where, a = 10, r = 102/10= 10, n = n

a(rn –1 )/(r – 1) =

(ii) 7 + 77 + 777 + … to nterms.

Let us take 7 as acommon term so we get,

7 [1 + 11 + 111 + … ton terms]

Now multiply and divideby 9 we get,

7/9 [9 + 99 + 999 + …n terms]

7/9 [(10 – 1) + (102 –1) + (103 – 1) + … + (10n – 1)]

7/9 [(10 + 102 +103 + … +10n)] – 7/9 [(1 + 1 + 1 + … to n terms)]

So the terms are inG.P

Where, a = 10, r = 102/10= 10, n = n

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

7/9 [10 (10n –1)/(10-1)] – n

7/9 [10/9 (10n –1) – n]

7/81 [10 (10n –1) – n]

7/81 (10n+1 –9n – 10)

(iii) 9 + 99 + 999 + … to nterms.

The given terms can bewritten as

(10 – 1) + (100 – 1) +(1000 – 1) + … + n terms

(10 + 102 +103 + … n terms) – n

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

Where, a = 10, r = 10,n = n

a(rn –1 )/(r – 1) = [10 (10n – 1)/(10-1)] – n

= 10/9 (10n –1) – n

= 1/9 [10n+1 –10 – 9n]

= 1/9 [10n+1 –9n – 10]

(iv) 0.5 + 0.55 + 0.555 +…. to n terms

Let us take 5 as acommon term so we get,

5(0.1 + 0.11 + 0.111 +…n terms)

Now multiply anddivide by 9 we get,

5/9 [0.9 + 0.99 +0.999 + …+ to n terms]

5/9 [9/10 + 9/100 +9/1000 + … + n terms]

This can be written as

5/9 [(1 – 1/10) + (1 –1/100) + (1 – 1/1000) + … + n terms]

5/9 [n – {1/10 + 1/102 +1/103 + … + n terms}]

5/9 [n – 1/10{1-(1/10)n}/{1 – 1/10}]

5/9 [n – 1/9 (1 – 1/10n)]

(v) 0.6 + 0.66 + 0.666 +…. to n terms.

Let us take 6 as acommon term so we get,

6(0.1 + 0.11 + 0.111 +…n terms)

Now multiply anddivide by 9 we get,

6/9 [0.9 + 0.99 +0.999 + …+ n terms]

6/9 [9/10 + 9/100 +9/1000 + …+ n terms]

This can be written as

6/9 [(1 – 1/10) + (1 –1/100) + (1 – 1/1000) + … + n terms]

6/9 [n – {1/10 + 1/102 +1/103 + … + n terms}]

6/9 [n – 1/10{1-(1/10)n}/{1 – 1/10}]

6/9 [n – 1/9 (1 – 1/10n)]

Question 5 : How many terms of the G.P. 3, 3/2, ¾,… Be taken together to make 3069/512 ?

Given:

Sum of G.P = 3069/512

Where, a = 3, r =(3/2)/3 = 1/2, n = ?

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

3069/512 = 3 ((1/2)n –1)/ (1/2 – 1)

3069/512×3×2 = 1 –(1/2)n

3069/3072 – 1 = –(1/2)n

(3069 – 3072)/3072 = –(1/2)n

-3/3072 = – (1/2)n

1/1024 = (1/2)n

(1/2)10 =(1/2)n

10 = n

10 terms are requiredto make 3069/512

Question 6 : How many terms of the series 2 + 6 + 18 + ….Must be taken to make the sum equal to 728?

Given:

Sum of GP = 728

Where, a = 2, r = 6/2= 3, n = ?

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

728 = 2 (3n –1)/(3-1)

728 = 2 (3n –1)/2

728 = 3n –1

729 = 3n

36 = 3n

6 = n

6 terms are requiredto make a sum equal to 728

Question 7 : How many terms of the sequence √3, 3,3√3,… must be taken to make the sum 39+ 13√3 ?

Given:

Sum of GP = 39 +13√3

Where, a =√3, r = 3/√3= √3, n = ?

By using the formula,

Sum of GP for n terms =a(rn – 1 )/(r – 1)

39 + 13√3 = √3 (√3n –1)/ (√3 – 1)

(39 + 13√3) (√3 – 1) =√3 (√3n – 1)

Let us simplify weget,

39√3 – 39 + 13(3) –13√3 = √3 (√3n – 1)

39√3 – 39 + 39 – 13√3= √3 (√3n – 1)

39√3 – 39 + 39 – 13√3= √3n+1 – √3

26√3 + √3 = √3n+1

27√3 = √3n+1

√36 √3= √3n+1

6+1 = n + 1

7 = n + 1

7 – 1 = n

6 = n

6 terms are requiredto make a sum of 39 + 13√3

Question 8 : The sum of n terms of the G.P. 3, 6, 12, … is 381. Find the value of n.

Given:

Sum of GP = 381

Where, a = 3, r = 6/3= 2, n = ?

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

381 = 3 (2n –1)/ (2-1)

381 = 3 (2n –1)

381/3 = 2n –1

127 = 2n –1

127 + 1 = 2n

128 = 2n

27 = 2n

n = 7

value of n is 7

Question 9 : The common ratio of a G.P. is 3, and the last term is 486. If the sum of these terms be 728, find the first term.

Given:

Sum of GP = 728

Where, r = 3, a = ?

Firstly,

Tn =arn-1

486 = a3n-1

486 = a3n/3

486 (3) = a3n

1458 = a3n ….Equation (i)

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

728 = a (3n –1)/2

1456 = a3n –a … equation (2)

Subtracting equation(1) from (2) we get

1458 – 1456 = a.3n –a.3n + a

a = 2.

The first termis 2

Question 10 : The ratio of the sum of the first three terms is to that of the first 6 terms of a G.P. is 125 : 152. Find the common ratio.

Given:

Sum of G.P of 3 termsis 125

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

125 = a (rn –1)/(r-1)

125 = a (r3 –1)/ (r-1) … equation (1)

Now,

Sum of G.P of 6 termsis 152

By using the formula,

Sum of GP for n terms= a(rn – 1 )/(r – 1)

152 = a (rn –1)/(r-1)

152 = a (r6 –1)/ (r-1) … equation (2)

Let us divide equation(i) by (ii) we get,

125/152 = [a (r3 –1)/ (r-1)] / [a (r6 – 1)/ (r-1)]

125/152 = (r3 –1)/(r6 – 1)

125/152 = (r3 –1)/[(r3 – 1) (r3 + 1)]

125/152 = 1/(r3 +1)

125(r3 +1) = 152

125r3 +125 = 152

125r3 =152 – 125

125r3 =27

r3 =27/125

r3 = 33/53

r = 3/5

The common ratiois 3/5

krishan