A. 1024

B. 980

C. 1280

D. 320

T

Therefore, T

A. 200

B. 120

C. 250

D. 100

⇒

A. 2,3, 4,5

B. 1,2,4,8

C. -12,144,-1728,

D. -1,2,-4,8

ar

Dividing 2 by 1:

r

r = 2

putting r=2 in equation(1)

we get a=1.

So the required GP= 1,2,4,8,16…

A. 5

B. 7

C. 8

D. 10

(ar)

a=6, and r= (-2)

Solving for n:

6 (-2)

(-2)

n-1 = 6

n=7

A. 1/8

B. 2/5

C. 1/2

D. 1/4

Since the 1^{st} term is 64 and the 5^{th} term is 4, it is obvious that successive terms decrease in value. Therefore, r < 1. So, the sum of all terms is a/(1 – r) = 128. Solving, we get r = ½.**Q.6.** Find the sum of GP.: 1, 2,4,8, ... up to the 10th term.**Answer:** **Option A****Explanation:** Using the formula for summation of n terms of a GP:

S_{n} = a(r^{n} -1)/(r-1)

Putting the value of a=1, r =2 , n=10

We get the solution :

2^{10} - 1**Q.7.** Find the sum of an infinite GP 3,1,1/3,…. ?**Ans. Option B****Explanation:** The sum of an infinite GP is given by,

S_{n} = a / 1-r

a=3, r=1/3.

Putting the values in the equation :

3 / (1-1/3)= 9/2**Q.8.** How many terms of the following G. P.: 64, 32, 16, … has the sum 127 ½ ?**Answer: Option A****Explanation:** : Here a = 64, r =

Using the formula , we get

⇒ ......(given)

or

or

or

∴ n = 8**Q.9.** Find the sum up to n terms of the sequence: 0.7, 0.77, 0.777, …**Answer:** **Option B****Explanation:** Let S denote the sum, then

S = 0.7 + 0.77 + 0.777 +..... to n term

= 7(0.1 + 0.11 + 0.111 +..... to n term )

= (0.9 + 0.99 + 0.999 + .... to n terms)

= {(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + .... to n terms}

= {(1+1+1+....n terms) - (0.1 + 0.01 + 0.001 + ... to n terms)}

=

= (Since r< 1)

=

=

=**Q.10.** Express the following recurring decimal as an infinite G. P. and find its value in rational form: **Answer:** **Option A****Explanation:** = 0.333333.....

=0.3+0.03+0.003+0.0003+......

=

The above is an infinite G.P. with the first term a = and

Hence, by using the formula we get

Hence, the recurring decimal

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A. 2^{10}-1

B. 2^{10}

C. 216

D. 2^{10}/2

S

Putting the value of a=1, r =2 , n=10

We get the solution :

2

A. 9

B. 9/2

C. 12

D. 24

S

a=3, r=1/3.

Putting the values in the equation :

3 / (1-1/3)= 9/2

A. 8

B. 5

C. 12

D. 24

Using the formula , we get

⇒ ......(given)

or

or

or

∴ n = 8

A. 7^{n-1}

B.

C. 7^{n}/81

D. 7^{n+1}/10^{n}

S = 0.7 + 0.77 + 0.777 +..... to n term

= 7(0.1 + 0.11 + 0.111 +..... to n term )

= (0.9 + 0.99 + 0.999 + .... to n terms)

= {(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + .... to n terms}

= {(1+1+1+....n terms) - (0.1 + 0.01 + 0.001 + ... to n terms)}

=

= (Since r< 1)

=

=

=

A. 1/3

B. 1/5

C. 1/7

D. 1/9

=0.3+0.03+0.003+0.0003+......

=

The above is an infinite G.P. with the first term a = and

Hence, by using the formula we get

Hence, the recurring decimal