GEOMETRIC SERIES WORKSHEET

Question 1 :

Find the sum of the following geometric series.

1 + 3 + 9 + .......... to 10 terms

Question 2 :

Find the sum of 8 terms of the following geometric series.

1 + (-3) + 9 + (-27) +..........

Question 3 :

Find the sum of 30 terms of the following geometric series.

7 + 7 + 7 + 7 +..........

Question 4 :

Find the sum of terms of the following geometric series.

1 + 2 + 4 + 8 +.......... + 2048

Question 5 :

Find the sum of the following infinite geometric series.

3 + 1 + 1/3 +.......... 

Question 6 :

Find the sum of the following infinite geometric series.

-3 + (-12) + (-48) +.......... 

Question 7 :

A ball is dropped from a height of 6 m and on each bounce it bounces 2/3 of its previous height.

(i) What is the total length of the downward paths?

(ii) What is the total length of the upward paths?

(iii) How far does the ball travel till it stops bouncing?

1. Answer :

1 + 3 + 9 + .......... to 10 terms

a1 = 1

r = a2/a1 = 3/1 = 3 > 1

n = 10

Formula to find sum of n terms of a geometric series where r  1.

Sn = a1(rn - 1)/(r - 1)

Substitute a1 = 1 and r = 3.

S10 = 1(310 - 1)/(3 - 1)

= (59049 - 1)/2

= 59048/2

= 29524

2. Answer :

1 + (-3) + 9 + (-27) +.......... to 8 terms

a1 = 1

r = a2/a1 = -3/1 = -3 < 1

n = 10

Formula to find sum of n terms of a geometric series where r > 1.

Sn = a1(rn - 1)/(r - 1)

Substitute a1 = 1, r = -3 and n = 8.

S8 = 1[(-3)8 - 1]/(-3 - 1)

= (6561 - 1)/(-4)

= 6560/(-4)

= -1640

3. Answer :

7 + 7 + 7 + 7 +.......... to 30 terms

a1 = 7

r = a2/a1 = 7/7 = 1

n = 30

Formula to find sum of n terms of a geometric series where r = 1.

Sn = na1

Substitute a1 = 7 and r = 1.

S30 = 30(7)

= 210

4. Answer :

1 + 2 + 4 + 8 +.......... + 2048

a1 = 1

r = a2/a1 = 2/1 = 2 > 1

The value of n is not given. So, we have to find the value of n.

Let 2048 be the nth term.

an = 2048

a1rn-1 = 2048

Substitute a1 = 1 and r = 2.

1(2)n-1 = 2048

2n-1 = 211

n - 1 = 11

n = 12

Formula to find sum of n terms of a geometric series where r  1.

Sn = a1(rn - 1)/(r - 1)

Substitute a1 = 1, r = -3 and n = 12.

S12 = 1[212 - 1]/(2 - 1)

= (4096 - 1)/1

= 4095

5. Answer :

3 + 1 + 1/3 +.......... 

a1 = 3

r = a2/a1 = 1/3

The value of r (= 1/3) is in the interval -1 < r < 1.

The sum for the given infinite geometric series exists.

S = a1/(1 - r)

Substitute a1 = 3 and r = 1/3.

S = 3/(1 - 1/3)

= 3/(2/3)

= 3(3/2)

= 9/2

6. Answer :

-3 + (-12) + (-48) +.......... 

a1 = 3

r = a2/a1 = -12/(-3) = 4

The value of r (= 4) is not in the interval -1 < r < 1.

So, the sum for the given infinite geometric series does not exist.

7. Answer :

(i) Distance covered in the downward path :

= 6 + 4 + 8/3 + 16/9 + .............

a1 = 6

r = a2/a1 = 4/6 = 2/3

The value of r (= 2/3) is in the interval -1 < r < 1.

Formula to find sum of infinite geometric series :

S = a1/(1 - r)

Substitute a1 = 6 and r = 2/3.

= 6/(1 - 2/3)

= 6/(1/3)

= 6(3/1)

= 18 m

(ii)  Distance covered in the upward path :

= 4 + 8/3 + 16/9 + ............

S = a1/(1 - r)

Substitute a1 = 4 and r = 2/3.

S = 4/(1 - 2/3)

= 4/(1/3)

= 4(3/1)

= 12 m

(iii) Total distance covered :

= 18 + 12

= 30 m

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