Note:

Question 1: Find the sum of the following series to infinity:

i)

ii)

iii)

iv)

v)

Answer:

i)

Here

ii)

Here

iii)

iv)

Here

v)

Question 2: Prove that

Answer:

LHS =

RHS. Hence proved.

Question 3: Prove that

Answer:

Let

Subtracting the two we get

Now substituting back we get

RHS. Hence proved.

Question 4: If denotes the sum of the series and the sum of the series , prove that

Answer:

Given:

Also

Hence proved.

Question 5: Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is , and the difference between the third and fifth term is equal to .

Answer:

Given

Let the common ratio be

Also given

Let

When

When

When

When

Question 6: Express the recurring decimal as a rational number.

Answer:

We have

Question 7: Find the rational number whose decimal expansion is

Answer:

Question 8: Find the rational numbers having the following decimal expansions: i) ii) iii) iv)

Answer:

i)

ii)

iii)

iv)

Question 9: One side of an equilateral triangle is cm. The mid-points of its sides are joined to form another triangle whose mid-points, in tum, are joined to form still another triangle. The process is continued indefinitely. Find the sum of the (i) perimeters of all the triangles. (ii) areas of all triangles.

Answer:

We know, the length of the first triangle

Length of the side of the second triangle

Length of the side of the third triangle

Hence the series is

i) Sum of the perimeters of the triangles

cm

ii) Area of the triangle .

We know, if the base of an equilateral triangle is , then the area

the sum of the area of the triangles

Question 10: Find an infinite G.P. whose first term is and each term is the sum of all the terms which follow it.

Answer:

Let the first term be and the common ratio be

Therefore G.P. is

Given

Therefore G.P. is

Question 11: The sum of first two terms of an infinite G.P. is and each term is three times the sum of the succeeding terms. Find the G.P.

Answer:

Let the first term be and the common ratio be

Therefore G.P. is

Given

Therefore

Therefore G.P. is

Question 12: Show that in an infinite G.P. with common ratio , each term bears a constant ratio to the sum of all terms that follow it.

Answer:

Let the G.P. be

Therefore

Therefore

Similarly,

and

Therefore we see that each of the ratio is constant.

Question 13: If denotes the sum of an infinite G.P. and denotes the sum of the squares of its terms, then prove that the first term and common ratio are respectively and .

Answer:

Let the first term be and the common ratio be .

Given … … … … … i)

… … … … … ii)

Now

Substituting in i)