# Question-1

**Write the first five terms of each of the sequences. whose nth terms are a**

_{n}= n(n+2)**Solution:**

a

_{n}= n(n+2)

For n = 1, a_{1} = 1(1 + 2) = 3

For n = 2, a_{2} = 2(2 + 2) = 8

For n = 3, a_{3} = 3(3 + 2) = 15

For n = 4, a_{4} = 4(4 + 2) = 24

For n = 5, a_{5} = 5(5 + 2) = 35

Thus first five terms are 3, 8, 15, 24, 35.

# Question-2

**Write the first five terms of each of the sequences. whose nth terms are**

**a**

_{n}=**Solution:**

a

_{n}=

For n = 1, a

_{1}= =

For n = 2, a

_{2}= =

For n = 3, a

_{3}= =

For n = 4, a

_{4}= =

For n = 5, a

_{5}= =

Hence first 5 terms of the given sequence are

# Question-3

**Write the first five terms of each of the sequences. whose nth terms are a**

_{n}= 2^{n }**Solution:**

a

_{n}= 2

^{n}

For n = 1, a

_{1}= 2

^{1}= 2

For n = 2, a_{2} = 2^{2} = 4

For n = 3, a_{3} = 2^{3} = 8

For n = 4, a_{4} = 2^{4} = 16

For n = 5, a_{5} = 2^{5} = 32

Hence first five terms of the given sequence are 2, 4, 8, 16 and 32.

# Question-4

**Write the first five terms of each of the sequences. whose nth terms are a**

_{n}=**Solution:**

a

_{n}=Substituting n = 1, 2, 3, 4 and 5 we get

a_{1} = (2 x 1 - 3)/6 = -1/6

a_{2} = (2 x 2 - 3)/6 = (4 - 3)/6 = 1/6

a_{3} = (2 x 3 - 3)/6 = (6 - 3)/6 = 3/6 = Â½

a_{4} = (2 x 4 - 3)/6 = (8 - 3)/6 = 5/6

a_{5} = (2 x 5 - 3)/6 = (10 - 3)/6 = 7/6

Therefore the first term is â€“1/6, 1/6, Â½, 5/6 and 7/6.

# Question-5

**Write the first five terms of each of the sequences. whose nth terms are a**

_{n}= (-1)^{n-1}5^{n+1 }**Solution:**

a

_{n}= (-1)

^{n-1}5

^{n+1}

Substituting n = 1, 2, 3, 4 and 5 we get

a_{1} = (-1)^{1-1} 5^{1+1}= (-1)^{0} 5^{2} = 25

a_{2} = (-1)^{2-1} 5^{2+1}= (-1)^{1} 5^{3} = -125

a_{3} = (-1)^{3-1} 5^{3+1}= (-1)^{2} 5^{4} = 625

a_{4} = (-1)^{4-1} 5^{4+1}= (-1)^{3} 5^{5} = -3125

a_{5}= (-1)^{5-1} 5^{5+1}= (-1)^{4} 5^{6} = 15625

Therefore the first term is 25, -125, 625, -3125 and 15625.

# Question-6

**Write the first five terms of each of the sequences. whose nth terms are a**

_{n}= n(n^{2}+ 5)/4**Solution:**

a

_{n}= n(n

^{2}+ 5)/4

Substituting n = 1, 2, 3, 4 and 5 we get

a_{1} = 1(1^{2} + 5)/4 = 6/4 = 3/2

a_{2} = 2(2^{2} + 5)/4 = 2(4 + 5)/4 = 18/4 = 9/2

a_{3} = 3(3^{2} + 5)/4 = 3(9 + 5)/4 = 135/2

a_{4} = 4(4^{2} + 5)/4 = 4(16 + 5)/4 = 84/4 = 21

a_{5} = 5(5^{2} + 5)/4 = 5(25 + 5)/4 = 150/2 = 75

Therefore the first term is 3/2, 9/2, 135/2, 21 and 75/2.

# Question-7

**Find the indicated terms in each of the sequences. whose nth terms are a**

_{n}= 4n â€“ 3, a_{17}, a_{24 }**Solution:**

For n = 17, a

_{17}= 4 Ã— 17 â€“ 3 = 68 â€“ 3 = 65

For n = 24, a_{24} = 4 Ã— 24 â€“ 3 = 96 â€“ 3 = 93

Hence a_{17} = 65 and a_{24} = 93

# Question-8

**Find the indicated terms in each of the sequences. whose nth terms are a**

_{n}=**Solution:**

a

_{n}

**=**

for n = 7, a

_{7}=

**=**

Hence a

_{7}=

# Question-9

**Find the indicated terms in each of the sequences. whose nth terms are a**_{n}= (-1)^{n â€“ 1}n^{3}, a_{9 }**Solution:**

a

_{n}= (-1)

^{n â€“ 1}. n

^{3}

for n = 9, a

_{9}= (-1)

^{9 â€“ 1}. (9)

^{3 }= 9 Ã— 9 Ã— 9 = 729

Hence a

_{n}= 729

# Question-10

**Find the indicated terms in each of the sequences. whose nth terms are a**

_{n}=**Solution:**

a

_{n}

**=**

For N = 20, a_{20} = 20 = =

Hence a_{20} =

# Question-11

**a**

_{1}= 3**a**

_{n}= 3a_{n â€“ 1}+ 2, for all n > 1

**Solution:**

a

_{1}= 3

a_{n} = 3a_{n â€“ 1} + 2, for all n > 1

For n = 2, a_{2} = 3a_{n â€“ 1} + 2, for all n > 1

= 3a_{1} + 2 = 3 Ã— 3 + 2 = 11

For n = 3, a_{3} = 3a_{2} + 2 = 3 Ã— 11 + 2 = 35

For n = 4, a_{4} = 3a_{3} + 2 = 3 Ã— 35 + 2 = 107

For n = 5, a_{5} = 3a_{4} + 2 = 3 Ã— 107 + 2 = 323

Hence the required sequence is

3, 11, 35, 107, 323+â€¦â€¦..

Corresponding series

= 3 + 11 + 35 + 107 + 323+â€¦â€¦..

# Question-12

**Write the first five terms of each of the sequence in obtain the corresponding series a**

_{1}= -1, a_{n}= a_{n-1}/n, (n â‰¥ 2)**Solution:**

a

_{1}= -1

a

_{n}=

For n = 2, a

_{2}=

For n = 3, a

_{3}= = -

For n = 4, a

_{4}= = -

For n = 5, a

_{5}= = -

Hence first 5 terms are

-1, -

and the series is

(-1) +

= -1 -

# Question-13

**Write the first five terms of each of the sequence in obtain the corresponding series a**

_{1}= a_{2}= 2, a_{n}= a_{n-1}â€“ 1, (n > 2)**Solution:**

a

_{1}= a

_{2}= 2

and a

_{n}= a

_{n â€“ 1, }n > 2

For n = 3, a

_{3}= a

_{2}â€“ 1 = 2 â€“ 1 = 1

For n = 4, a

_{4}= a

_{3}â€“ 1 = 1 â€“ 1 = 0

For n = 5, a

_{5}= a

_{4}â€“ 1 = 0 â€“ 1 = -1

Hence first 5 terms are 2, 2, 1, 0, -1 and the corresponding series = 2 + 2 + 1 + 0 +(-1)+â€¦

# Question-14

**The Fibonacci sequence is defined by a**

Find a

_{1}= 1 = a_{2}, a_{n}= a_{n-1 }+ a_{n-2}^{ }(n > 2).Find a

_{n+1}/a_{n}, for n = 1, 2, 3, 4, 5.**Solution:**

Substituting n = 3, 4, 5 and 6 in a

_{n}= a

_{n-1 }+ a

_{n-2 }

a

_{3}= a

_{3-1 }+ a

_{3-2 }= a

_{2 }+ a

_{1}= 1 + 1 = 2

a

_{4}= a

_{4-1 }+ a

_{4-2 }= a

_{3 }+ a

_{2}= 2 + 1 = 3

a

_{5}= a

_{5-1 }+ a

_{5-2 }= a

_{4 }+ a

_{3}= 3 + 2 = 5

a

_{6}= a

_{6-1 }+ a

_{6-2 }= a

_{5 }+ a

_{4}= 5 + 3 = 8

Substituting n = 1

Substituting n = 2, 3, 4 and 5 we get

** **

The required series is 1, 2, 3/2, 5/3 and 8/5.

# Question-15

**Find the sum of odd integers from 1 to 2001.**

**Solution:**

S

_{n}= 1 + 3 + 5 + 7 + â€¦.. + 2001

Now a = 1

d = 3 â€“ 1 = 2, l = 2001

Let t_{n} = 2001

Therefore a + (n + 1) d = 2001

or 1 + (n â€“ 1)(2) = 2001

or (2n â€“ 1) = 2000

or n â€“ 1 = 1000

or n = 1001

We have S_{n} =

or S_{1001} =

= = 1001 Ã— 1001

= 1002001

# Question-16

**Find the sum of all natural numbers between 100 and 1000 which are multiples of 5.**

**Solution:**

The next number to 100 which is multiple of 5 is 105. Similarly the next number to 105 multiple of 5 is 110. The greatest number multiple of 5 but less than 1000 is 995.

a = 105,

d = 110 â€“ 105 = 5

a_{n }= a + (n - 1)d

995 = 105 + (n - 1)5

995 = 105 + 5n â€“ 5

5n = 895

n = 179

Son =

=

= 98450

Therefore the sum of all natural numbers between 100 and 1000 are 98450.

# Question-17

**In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the sum of the next five terms. Show that the 20**

^{th}term is â€“112.**Solution:**

a = 2, S

_{5}=

_{ }(1/4) (S

_{10 }- S

_{5})

S_{5}= (5/2)(2ï‚´ 2 + 4d) = (5/2)(4 + 4d) = 10 + 10d

S_{10 }= (10/2)(2ï‚´ 2 + 9d) = (10/2)(4 + 9d) = 5(4 + 9d) = 20 + 45d

Since S_{5} =_{ }(1/4) (S_{10 }- S_{5})

10 + 10d = (1/4)(20 + 45d â€“ 10 - 10d)

40 + 40d = 10 + 35d

-5d = 30

d = -6

a_{20 }= 2 + (20 â€“1)(-6)

= 2 + (20 â€“1)(-6)

= 2 â€“ 114

= - 112

# Question-18

**How many terms of the A.P. â€“6, -11/2, -5, â€¦.. are needed to give the sum â€“25?**

**Solution:**

a= -6, d = (-11/2) + 6 = (-11 + 12)/2 = Â½, S

_{n}= -25

-25 = (n/2)[2(-6) + (n -1)(1/2)]

-25ï‚´ 4 = n(-24 + n â€“1)

- 100 = n(-25 + n)

n^{2} -25n + 100 = 0

n^{2} -20n â€“ 5n + 100 = 0

n(n â€“ 20) â€“ 5(n â€“ 20) = 0

(n â€“ 5)(n â€“ 20) = 0

n = 5, n = 20

Therefore the required number of terms is 5 or 20.

# Question-19

**If the p**

^{th}term of an A.P is and the q^{th}term is , prove that the sum of the first pq terms must be (pq + 1).**Solution:**

t

_{p = }a + (p-1)d = â€¦â€¦â€¦.(1)

t_{q} = a + (q-1)d = â€¦â€¦â€¦.(2)

(1) - (2)

a + (p-1)d =

a + (q-1)d =

---------------------

(p - q)d =

----------------------

d = =

a + (p-1) =

a = - = =

Sum of the 1^{st} pq terms = ]

= =

S_{pq} =

# Question-20

**If the sum of a certain number of terms of the A.P. 25, 22, 19,â€¦.. is 116, find the last term.**

**Solution:**

Given. A.P.: 25, 22, 19,â€¦..

Here Solution = 116, a = 25, d = 22 â€“ 25 = -3

Let n be the number of terms.

i.e., Last term = t_{n }

Now S_{n} =

or 116 =

or 232 = n[50 â€“ 3n + 3]

or n[53 â€“ 3n] = 232

or 3n^{2} â€“ 53n + 232 = 0

or 3n^{2} â€“ 29n â€“ 24n + 232 = 0

or n(3n â€“ 29) â€“8(3n â€“ 29) =0

or (3n â€“ 29) (n â€“ 8) =0

or n = 8 as n âˆ‘ N

Now Last term = T_{n} = a+(n â€“ 1) d

or T_{8} = 25 + (8 â€“ 1)(-3)

= 25 â€“ 21 = 4

Therefore the last term is 4.

# Question-21

**Find the sum to n terms of the A.P. whose kth term is 5k + 1.**

**Solution:**

T

_{k}= 5k + 1

T_{1} = 5 Ã— 1 + 1 = 6

T_{2} = 5 Ã— 2 + 1 = 11

T_{3} = 5 Ã— 3 + 1 = 16

T_{4} = 5 Ã— 4 + 1 = 21

Now a = 6, d = 11 â€“ 6 = 5

Thus S_{n} =

** = **

** **=

** **= =

# Question-22

**If the sum of n terms of an A.P is (pn + qn**

^{2}), where p and q are constants, find the common difference.**Solution:**

For the A.P.

S_{n} = pn + qn^{2}

Now S_{1} = p Ã— 1 + q(1)^{2 }

S_{1} = p Ã— q â‡’ T_{1} = p + q

Also S_{2} = p Ã— 2 + q(2)^{2}

= 2p + 4q

We have T_{1} + T_{2} = 2p + 4q

or T_{2} = 2p + 4q â€“ T_{1 }

or T_{2} = 2p + 4q â€“ (p + q)

or T_{2} = p + 3q

Hence common difference = T_{2} â€“ T_{1}

= p + 3q â€“(p + q)

= p + 3q â€“ p â€“ q = 2q

# Question-23

**If the sum of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6, find the ratio of their 18**

^{th}terms.**Solution:**

For first A.P. :

Let first term = a

_{1 }

Common difference = d_{1 }

and S_{n1} = Sum of n terms

For second A.P. :

Let first term = a_{2 }

Common difference = d_{2 }

and S_{n2} = Sum of n terms

Given =

Required to find

=

We have =

or =

or =

Taking = 17 or n = 34 + 1 = 35, we have = =

# Question-24

**If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first(p+q) terms.**

**Solution:**

Given. For the A.P.

S_{p} = S_{q}

Let the first term = a, common difference = d

Therefore, according to the question** ** =

Required. To find S_{p + q }

Now =

According to question

or p[2a + pd â€“ d] = q[2a + qd â€“ d]

or 2ap + p^{2}d â€“ pd = 2aq + q^{2}d â€“ qd

or 2a(p - q)^{2 }+d(p^{2} â€“ q^{2}) â€“ d(p â€“q) = 0

or 2a + d(p + q) - d = 0

2a + (p + q â€“ 1)d = 0

Now S_{p + q} =

= = 0

# Question-25

**Sums of the first p, q, r terms of an A.P are a, b, c respectively. Prove that**

**Solution:**

Let x be the first term and d the common difference.

S_{p} = p/2[ 2x + (p-1)d] = a

S_{q} = q/2[2x + (q-1)d] = b

S_{r} = r/2[2x + (r-1)d] = c

2x + (p-1)d = 2a/pâ€¦â€¦â€¦..(i)

2x + (q-1)d = 2b/qâ€¦â€¦â€¦..(ii)

2x + (r-1)d = 2c/râ€¦â€¦â€¦â€¦.(iii)

Eliminating x and d from (i), (ii) and (iii).

(p-q)d = 2a/p - 2b/q [Subtracting (i) and (ii)] â€¦â€¦(iv)

(q-r)d = 2a/q - 2c/r [Subtracting (iii) and (ii)] â€¦ (v)

[ Dividing (iv) and (v)]

=

# Question-26

**The ratio of the sums of m and n terms of an A.P is m**

^{2}:n^{2}. Show that the ratio of m^{th}and n^{th}term is 2m â€“ 1: 2n â€“ 1.**Solution:**

S

_{m}: S

_{n}= m

^{2}:n

^{2 }

S_{m} : m^{2 }= S_{n }: n^{2} = k

T_{m} : T_{n} = S_{m} â€“ S_{m - 1 }: S_{n} â€“ S_{n â€“ 1 }

= k{m^{2} â€“ (m - 1)^{2}} : k{n^{2} â€“ (n - 1)^{2}}

= 2m â€“ 1 : 2n - 1

# Question-27

**If the sum of n terms of an A.P. is 3n**

^{2}+ 5n and its mth term is 164, find the value of m.**Solution:**

Given. S

_{n}= 3n

^{2}+ 5n

S_{1} = 3(1)^{2} + 5 Ã— 1 = 8 â‡’ T_{1} = 8

S_{2} = 3(2)^{2} + 5 Ã— 2 = 22

Thus T_{1} + T_{2} = 22

or 8 + T_{2} = 22

or T_{2} = 14

Now a = 8, d = 14 â€“ 8 = 6

As T_{m} = 164

or a +(m â€“ 1)d = 164

or 8 + (m â€“ 1) (6) = 164

or 8 + 6m â€“ 6 = 164

or 6m = 162

or m = = 27

Hence m = 27

# Question-28

**Insert five number between 8 and 26 such that the resulting sequence is an A.P.**

**Solution:**

Let the five A.Mâ€™s be A

_{1}, A

_{2}, A

_{3}, A

_{4}, A

_{5}

Therefore 8, A_{1}, A_{2}, A_{3}, A_{4}, A_{5, }26 are in A.P. Let d be the common difference

Here T_{1} = 8

T_{7} = 26

â‡’a + ( 7 â€“ 1)d = 26

or a + 6d = 26

or 8 + 6d = 26

or 6d = 18

or d = 3

Therefore A_{1} = 8 + 3 = 11

A_{2} = 11 + 3 = 14

A_{3} = 14 + 3 = 17

A_{4} = 17 + 3 = 20

A_{5} = 20 + 3 = 23

Thus five required numbers are

11, 14, 17, 20 and 23

# Question-29

**If (a**

^{n}+ b^{n})/(a^{n-1}+ b^{n-1}) is the A.M. between a and b, then find the value of n.**Solution:**

The A.M between a and b is (a + b)/2.

(a

^{n}+ b

^{n})/(a

^{n-1}+ b

^{n-1}) = (a + b)/2

2(a^{n} + b^{n}) = (a + b)(a^{n-1} + b^{n-1})

2a^{n} + 2b^{n} = a^{n} + b^{n} + ba^{n-1 }+ ab^{n-1 }

a^{n} + b^{n} - ba^{n-1 }- ab^{n-1} = 0

a^{n} - ba^{n-1 }+ b^{n} - ab^{n-1} = 0

a^{n-1}(a â€“ b)^{ }- b^{n-1}(b â€“ a) = 0

(a^{n-1 }- b^{n-1})(b â€“ a) = 0

(a^{n-1 }- b^{n-1}) = 0 or (b â€“ a) = 0

a^{n-1 }= b^{n-1 }or b = a

If a = b, then n can take any value.

If a = b, then a^{n-1 }= b^{n-1 = }a^{n-1}/b^{n-1 }= 1 = (a/b)^{n-1 }= 1

As a = b, a/b = 1.

Therefore, (a/b)^{n-1 }= 1 = n â€“ 1 = 0 = n = 1

# Question-30

**Between 1 and 31, m arithmetic means have been inserted in such a way that the ratio of the 7**

^{th}and (m - 1)^{th}means is 5:9. Find the value of m.**Solution:**

Let the means be x

_{1}, x

_{2}, x

_{3}, x

_{4},â€¦â€¦, x

_{m}.

So that 1, x_{1}, x_{2}, x_{3}, x_{4},â€¦â€¦, x_{m}, 31 is an A.P of m + 2 terms

a_{m + 2} = a + (m + 1)d

31 = 1 + (m + 1)d

(m + 1)d = 30

d = 30/(m + 1) â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)

x_{7} : x_{m - 1} = 5:9

a + 7d : a + (m - 1)d = 5 : 9

9(1 + 7d) = 5(1 + (m - 1)d)

9 + 63d = 5 + 5md - 5d

4 = -63d + 5md - 5d

4 = (-68 + 5m)d

4 = (-68 + 5m)[30/ (m +1)]

4(m + 1) = (-68 + 5m)30

2m + 2= -1020 + 75m

73m = 1022

m = 14

# Question-31

**A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs.5 every month, what amount will he pay in the 30**

^{th}instalment.**Solution:**

Thus he pays Rs.100, Rs.105, Rs.100â€¦â€¦

Now a = 100, d = 5

To find, T_{30}

We know that

T_{n} = a + (n - 1)d

T_{30} = 100 + (30 - 1) Ã— 5

= 100 + 145

= 245

# Question-32

**The difference between any two consecutive interior angles of a polygon is 5**

^{0}. If the smallest angle is 120^{0}, find the number of the sides of the polygon.**Solution:**

We know that if the number of sides is n, then the sum of interior angles of a polygon with

n sides = (2n - 4) right angles

= (180n- 360) degrees

Now the sequence of angles

120Â°, 125Â°, 130Â°,... form an A.P. with a = 120Â°, d = 5Â°

Sn =

â‡’ 180n â€“ 360 =

or 2(180n â€“ 360) = n[240 + 5n - 5)]

or 360n - 720 = 240n + 5n^{2} + 5n

or 5n^{2 }â€“ 125n + 720 = 0

or n^{2} - 25n + 144 =0

or (n - 9)(n - 16) = 0

or n = 9, 16

But n = 16 is not possible as it gives the last term of A.P.

= a + (n - 1)d

= 120 + (16-9) x 5

= 195Â°

Thus the number of sides = 9.

# Question-33

**Find the 20**

^{th}and nth terms of the G.P.

**Solution:**

G.P is

Here, a = r = =

T_{n} = ar^{n â€“ 1}

Thus T_{20} = = =

and T_{n} =

# Question-34

**Find the 12**

^{th}term of a G.P. whose 8^{th}term is 192 and the common ratio is 2.**Solution:**

r = 2

a_{8} = ar^{7}

192 = a Ã— 2^{7}

192/2^{7} = a

a_{12} = ar^{11} = = 192 Ã— 2^{4} = 192 Ã— 16 = 3072

# Question-35

**The 5**

^{th}, 8^{th}and 11^{th}terms of a G.P. are p, q and s, respectively. Show that q^{2}= ps.**Solution:**

Let a and r be the first term and common ratio respectively.

5

^{th}term = a

_{5}= ar

^{4}

p = ar

^{4 }

a_{8} = ar^{7 }

q = ar^{7 }

a_{11} = ar^{10 }

s = ar^{10}

q^{2} = (ar^{7})^{2 }= a^{2} r^{14 }

ps = ar^{4}** **ar^{10 }= a^{2}r^{4+10 }= a^{2} r^{14 }

âˆ´ q^{2} = ps

# Question-36

**The 4**

^{th}term of a G.P. is square of its 2^{nd}term, and the first term is â€“3. Determine its 7^{th }term.**Solution:**

Let a and r be the first term and common ratio respectively.

Then a = -3

4^{th} term = a_{4} = ar^{3} = -3r^{3 }

2^{th} term =^{ }a_{2} = ar = -3r

a_{4 }= (a_{2})^{2 }

-3r^{3}_{ }= (-3r)^{2 }

-3r^{3}_{ }= 9r^{2 }

r_{ }= -3

7^{th} term = a_{7} = ar^{6} = -3(-3)^{6 }= -3(-3)^{6}= (-3)^{7} = - 2187

# Question-37

**(a) a = , r = 3/ =**

(b) a = , r =

(b) a = , r =

**Solution:**

(a) a = , r = 3/ =

Let 729 be the nth term of the G.P.

a_{n} = ar^{n-1 }

729 = ()^{n-1}

()^{12} = ()^{n}

n = 12

Therefore 729 is the 12^{th} term of the G.P.

(b) a = , r =

Let be the nth term of the G.P.

a_{n} = ar^{n-1
}**= **

**= **

n = 9

Therefore 1/19683 is the 9^{th} term of the G.P.

# Question-38

**For what values of x, the numbers , x, are in G.P?**

**Solution:**

x

^{2}= Ã— = 1

x = Â± 1

Therefore the required values of x are Â± 1.

# Question-39

**Find the sum of indicated terms of the following geometric progressions. 0.15, 0.015, 0.0015, â€¦..; 20 terms.**

**Solution:**

a = 0.15, r = 0.015/0.15 = 0.1 < 1

Therefore S

_{n}= = = = =

Substituting n = 20

S

_{20}=

# Question-40

**Find the sum of indicated terms of the following geometric progressions. , , 3, â€¦.; n terms.**

**Solution:**

a

**= ,**r

**= >**1

Therefore S

_{n}

**=**

** = **

** = **

** = **

** =**

# Question-41

**Find the sum of indicated terms of the following geometric progressions. 1, -a, a**

^{2}, -a^{3}, â€¦.; n terms. (a ï‚¹ -1)**Solution:**

a = 1, r = -a < 1

Therefore S

_{n}= = =

# Question-42

**Find the sum of indicated terms of the following geometric progressions. x**(x â‰ Â±1)

^{3}, x^{5}, x^{7}, â€¦.; n terms.**Solution:**

a = x

^{3}, r = x

^{2}< 1

Therefore S

_{n}= =

# Question-43

**Evaluate****.**

**Solution:**

= (2 + 3) + (2 + 3

^{2}) + (2 + 3

^{3}) + â€¦.. + (2 + 3

^{11}).

= (2 + 2 + â€¦â€¦â€¦â€¦.. 11 times) + 3 + 3^{2} + 3^{3} + â€¦.. + 3^{11}.

a = 3, r = 3^{2}/3 = 3 > 1

S_{n} = = =

Substituting n = 11

S_{11}^{ }=

Therefore the required sum is 22 + .

# Question-44

**The sum of first three terms of a G.P. is**

**and their product is 1. Find the common ratio and the terms.**

**Solution:**

Let three terms in G.P. are

** **

Their product = ** = **1

or a^{3} = 1

or a = 1

Again ** = **

or ** = **

or ** = **r

or 1 + 10r + 10r^{2} = 39r

or 10r^{2} - 29r + 10

or 10r^{2} - 25r â€“ 4r + 10 = 0

or 5r(2r â€“ 5) â€“ 2(2r â€“ 5) = 0

or (2r â€“ 5) (5r â€“ 5) = 0

or r =

Case (i) When r = ** **

Numbers are

Case (ii) When r = ** **

Numbers are

or or

Hence three numbers are or

# Question-45

**How many terms of the G.P. 3, 3**

^{2}, 3^{3}, â€¦â€¦ are needed to give the sum 120?**Solution:**

Given : 3 + 3

^{2}+ 3

^{3}+ â€¦.. = 120

Now a = 3, r = 3. Let number of terms = n

Also = 120

or = 120

or 3^{n} â€“ 1 =

or 3^{n} â€“ 1 = 80

or 3^{n} = 81 = 3^{4}

Hence n = 4

# Question-46

**The sum of the first three terms of a G.P is 16 and the sum of the next three terms is 128. Determine the first term, common ratio and the sum to n terms of the G.P.**

**Solution:**

Let the three terms be a, ar, ar

^{2}.

a + ar + ar^{2} = 16 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

Let the next three terms be ar^{3}, ar^{4}, ar^{5}.

ar^{3} + ar^{4} + ar^{5} = 128

r^{3}(a + ar + ar^{2}) = 128

r^{3}(16) = 128 (from i)

r^{3} = 8

r = 2

Therefore the common ratio is 2.

Substituting r = 2 in (i)

a + 2a + 4a = 16

7a = 16

a = 16/7

S_{n} =

Therefore the first term is 16/7, common ratio is 2 and sum to n terms is.

# Question-47

**Given a G.P. with a = 729 and 7**

^{th}term 64, determine S_{7}.

**Solution:**

a = 729

7

^{th}term = 64

âˆ´ a

_{7}= ar

^{6}, where r is the common ratio.

729r^{6} = 64

r^{6} = 64/729

r^{6} = (2/3)^{6 }

r = 2/3 < 1

S_{7} =

=

=

=

=

=

= 2059

âˆ´ S_{7} = 2059

# Question-48

**Find a G.P. for which sum of first two terms is -4 and the fifth term is 4 times the third term.**

**Solution:**

a + ar = -4

a(1 + r)= -4 â€¦â€¦â€¦..(1)

ar

^{4}= 4ar

^{2}

r

^{2}= 4

r = Â±2

Substituting r = 2 in (1)

a(1+2) = -4

a =

Substituting r = -2 in (2)

a(1-2) = -4

a = 4

.

^{.}. If a = ,r=2

or a = 4, r = -2.

# Question-49

**If the 4**

^{th}, 10^{th}and 16^{th}terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.**Solution:**

a

_{4}= ar

^{3}

x = ar

^{3}â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

a_{10} = ar^{9 }

y = ar^{9}â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

a_{16} = ar^{15 }

z = ar^{15}â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iii)

(ii) ï‚¸ (i)

y/x = ar^{9}/ar^{3} = r^{6} â€¦â€¦â€¦â€¦â€¦â€¦â€¦(iv)

(iii) ï‚¸ (ii)

z/y = ar^{15}/ar^{9} = r^{6 }â€¦â€¦â€¦â€¦â€¦â€¦â€¦(v)

From (iv) and (v)

y/x = z/y

y^{2} = xz.

âˆ´ x, y, z are in G.P

# Question-50

**Find the sum to n terms of the sequence 8,88,888,888,â€¦..**

**Solution:**

S_{n} = 8 + 88 + 888 + 8888 + â€¦â€¦ n terms

S_{n} = 8[1 +11 +111 + 1111 +â€¦â€¦. n terms]

=

=

=

=

=

# Question-51

**Find the sum of the products of the corresponding terms of the sequence 2, 4, 8, 16, 32 and 128, 32, 8, 2, .**

**Solution:**

First sequence : 2, 4, 8, 16, 32

Second sequence : 128, 32, 8, 2, Â½

New sequence after finding the products of the corresponding terms

2 Ã— 128, 4 Ã— 32, 8 Ã— 8, 16 Ã— 2, 32 Ã—

or 256, 128, 64, 32, 16

Now a = 256 r = = n = 5

Now S_{n} =

S_{n} =

=

= 512 Ã— = 496

# Question-52

**Show that the products of the corresponding, terms of the sequences a, ar, ar**

^{2}, â€¦.ar^{n â€“ 1}and A, AR, AR^{2},â€¦â€¦AR^{n â€“ 1}form a G.P. and find the common ratio.**Solution:**

Ist G.P :a, ar, ar

^{2}, â€¦â€¦.ar

^{n â€“ 1}

2^{nd} G.P. : A, AR, AR^{2},â€¦.. AR^{n â€“ 1 }

The sequence formed after multiplying the corresponding terms of the sequences is

(aA), (aA),(rR), (aR)r^{2}R^{2},â€¦..(aA)r^{n â€“ 1} R^{n â€“ 1}

Here = = rR

= = rR

= = rR

Since the ratios of two succeeding terms are the same, the resulting sequence is also a G.P.

The common ratio of the new G.P. = (rR).

# Question-53

**Find four numbers forming a Geometric progression in which the third term is greater than the first by 9, and the second term is greater than the fourth by 18.**

**Solution:**

The four terms of G.P are a, ar, ar

^{2}, ar

^{3}.

ar

^{2}- a = 9

a (r

^{2}-1) = 9 â€¦â€¦â€¦..(1)

ar (1 - r

^{2}) = 18 â€¦â€¦..(2)

Dividing (2) by (1)

r = -2

Substitute r in (1)

a (4-1) = 9

a = 3

.

^{.}. the four numbers are 3, 3(-2), 3(-2)

^{2}and 3(-2)

^{3}

(i.e) 3, -6, 12, -24.

# Question-54

**If the p**

^{th}, q^{th}and r^{th}terms of a G.P. are a, b, c respectively, prove that a^{q-r}. b^{r-p}. c^{p-q}= 1.**Solution:**

Let x be the first term and R the common ratio.

T

_{p}= x R

^{p-1}= a â€¦â€¦â€¦(1)

T

_{q}= x R

^{q-1}= b â€¦â€¦â€¦.(2)

T

_{r}= x R

^{r-1}= c â€¦â€¦â€¦(3)

a

^{q-r}

**= =**...........(1)

b

^{r-p}

**=**â€¦â€¦â€¦â€¦.(2)

c

^{p-q}

**=**â€¦â€¦â€¦.(3)

L.H.S. = (1) x (2) x (3)

.=1.

# Question-55

**If the first and the n**

^{th}terms of a G.P are a and b respectively and if p is the product of the first n terms, prove that P^{2}= (ab)^{n}.**Solution:**

Let the G.P be c, cr, cr

^{2}, â€¦â€¦â€¦.cr

^{n}

1

^{st}term c = a

n

^{th}term cr

^{n-1}= b

If n terms is c, cr, cr

^{2}, â€¦â€¦â€¦.cr

^{n-1}

Product of n term

p = c.(cr)(cr

^{2})(cr

^{3})â€¦â€¦..(cr

^{n-1})

= c

^{n}.r

^{1+2+â€¦â€¦â€¦n-1}

= c

^{n}

P

^{2}= (c

^{n})

^{2 2}= c

^{2n}r

^{n(n-1)}= c

^{2n}

_{r}n

^{2-n}â€¦â€¦â€¦â€¦.(1)

_{(ab)}n

_{= (c.cr}n-1)n

_{= (c}2

_{r}n-1

_{)}n

_{= c}2n

_{r}n

^{2-n}â€¦â€¦â€¦.(2)

From (1) and (2),

P

^{2}= (ab)

^{n}.

# Question-56

**Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n+1)th to (2n)th term is 1/r**

^{n}.**Solution:**

Consider the G.P. a, ar, ar

^{2},â€¦â€¦.

S_{n} =

Sum of terms from (n + 1)th to (2n)th terms

= S_{2n} - S_{n }

=

Now Required ratio =

=

=

=

=

= = R.H.S.

# Question-57

**If a, b, c, d are in G.P., show that (a**

^{2}+ b^{2}+ c^{2})(b^{2}+ c^{2}+ d^{2}) = (ab + bc + cd)^{2 }**Solution:**

Since a, b, c, d are in G.P

b/a = c/b = d/c = r, where r is the common ratio.

âˆ´ b = ar, c = br = ar

^{2}and d = cr = ar

^{3}.

L. H. S = (a

^{2}+ b

^{2}+ c

^{2})(b

^{2}+ c

^{2}+ d

^{2})

= [a

^{2}+ (ar)

^{2}+ (ar

^{2})

^{2}][( ar)

^{2}+ (ar

^{2})

^{2}+ (ar

^{3})

^{2}]

= (a^{2} + a^{2}r^{2} + a^{2}r^{4 })(a^{2}r^{2} + a^{2}r^{4} + a^{2}r^{6})

= a^{4} r^{2} (1 + r^{2} + r^{4 })(1 + r^{2} + r^{4})

= a^{4} r^{2} (1 + r^{2} + r^{4 })^{2}

R.H.S = (ab + bc + cd)^{2
}= (aï‚´ ar + ar ï‚´ ar^{2} + ar^{2 } ï‚´ï€ ar^{3})^{2 }

= (a^{2}r + a^{2}r^{3} + a^{2}r^{5})^{2}

= (a^{2}r)^{2}(1 + r^{2} + r^{3})^{2}

= a^{4} r^{2} (1 + r^{2} + r^{4 })^{2}

L.H.S = R. H. S

# Question-58

**Insert two numbers between 3 and 81 so that the resulting sequence is G.P.**

**Solution:**

Let the two numbers between 3 and 81 be G

_{1 }and G

_{2}.

Thus 3, G_{1}, G_{2}, 81 are in G.P.

Let r be the common ratio.

Therefore ar^{3} = 81

or 3r^{3} = 81

or r^{3} = 27 or r = 3

or G_{1} = 3 Ã— 3 = 9

G_{2} = 9 Ã— 3 = 27

Hence the two required numbers are 9 and 27.

# Question-59

**Find the value of n so that (a**

^{n+1 }+ b^{n+1})/(a^{n}+ b^{n}) may be the geometric mean between the a and b.**Solution:**

G

_{n }=

(a

^{n+1 }+ b

^{n+1})/(a

^{n}+ b

^{n}) =

a^{n+1 }+ b^{n+1} = a^{1/2}b^{1/2}(a^{n} + b^{n})

a^{n+1 }+ b^{n+1} = a^{n + (Â½) }b^{1/2} + a^{1/2}b^{n + (Â½) }

a^{n+1 }- a^{n + (Â½) }b^{1/2} = a^{1/2}b^{n + (Â½)} - b^{n+1 }

a^{n + (Â½) }(a^{Â½ }- b^{Â½}) = b^{n + (Â½) }(a^{Â½ }- b^{Â½})

a^{n + (Â½) }= b^{n + (Â½)}

a^{n}a^{(Â½) }= b^{n }b^{(Â½) }

(a/b)^{n }= (a/b)^{-Â½ }

n = -Â½

Hence value of n is -Â½.

# Question-60

**The sum of two numbers is 6 times their geometric mean. Show that the numbers are in the ratio 3 + 2 ïƒ–2 : 3 â€“ 2ïƒ–2.**

**Solution:**

Let the two numbers be a and b.

Then a + b = 6

(a + b)^{2} = 36ab

a^{2} + b^{2} + 2ab = 36ab

^{}

^{ }

^{}

^{ }

^{}

^{ }

# Question-61

**If A and G be A.M. and G.M. respectively between two positive numbers, prove that the numbers are A Â± .**

**Solution:**

Let the two positive numbers be a and b.

Therefore, A = â‡’ a + b = 2A

and G = â‡’ ab = G^{2}

The equation whose roots are a and b is

x^{2} â€“ (a + b)x + ab = 0

â‡’x^{2} â€“ 2Ax + G^{2} = 0 â€¦â€¦.(i)

or x =

= A Â±

= A Â±

The roots of (i) are a and b,

Hence, the two positive number are given by

A Â±

# Question-62

**The number of bacteria in a certain culture doubles very hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2**

^{nd}hour? 4th hour? nth hour?**Solution:**

As per question, the number of bacteria in a certain culture doubles every hour. So, they form a G.P. in which a = 30 and r = 2.

Bacteria present after 2^{nd} hour, 4^{th} hour and nth hour are T_{3}, T_{5} and T_{n+1} i.e., ar^{2}, ar^{4} and ar^{n}.

i.e., 30(2)^{2}, 30(2)^{4} and 30(2^{n}) respectively.

# Question-63

**What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?**

**Solution:**

T

_{1}= amount at the end of first year

= 500 = 500 Ã—

T_{2} = 500 Ã— â€¦..

Here a = 500 Ã— r =

n = 10

T_{n} = ar^{n â€“ 1 }

T_{10} = 500 Ã—

= Rs. 500

# Question-64

**If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively, then obtain the quadratic equation.**

**Solution:**

Let the quadratic equation be

(x - Î± )(x - Î² ) = 0

or x^{2} - (Î± + Î² )x + Î± Î² = 0

Roots are x = Î± , Î²

Therefore = 8 and = 5

or Î± + Î² = 8, and = 5

or = 25

Required quadratic equation is

x^{2} â€“ 8x + 25 = 0

# Question-65

**Find the sum of the following series upto n terms:1 Ã— 2 + 2 Ã— 3 + 3 Ã— 4 + 4 Ã— 5+â€¦â€¦**

**Solution:**

Let T

_{n}denotes the nth term of the given series

T_{n} = [nth term of 1, 2, 3 â€¦..] Ã— [nth term of 2, 3, 4 â€¦..]

= [1 + (n â€“ 1) 1] [2 + (n â€“ 1) 1]

= n(n + 1)

T_{n} = n^{2} + n

S_{n} = âˆ‘ n^{2} + âˆ‘ n

=

=

=

=

=

=

# Question-66

**Find the sum of the following series upto n terms 1 Ã— 2 Ã— 3 + 2 Ã— 3 Ã— 4 + 3 Ã— 4 Ã— 5 +â€¦â€¦.**

**Solution:**

Let T

_{n}denote the nth term of the given series .

Then,

T_{u} = [nth term of 1, 2, 3,â€¦.]

[nth term of 2, 3, 4,â€¦â€¦]

[nth term of 3, 4, 5,â€¦â€¦]

= [1 + (n â€“ 1)1] [2 + (n â€“ 1).1] [3 + (n â€“1).1]

= n(n + 1) (n + 2)

= n(n^{2} + 2n + 3) = n^{3} + 3n^{2} + 2n

âˆ´ S_{n} = âˆ‘ n^{3} + 3âˆ‘ n^{2 }+ 2âˆ‘ n

=

=

=

=

=

# Question-67

**Find the sum of the following series upto n terms 3 Ã— 1**

^{2}, 5 Ã— 2^{2}, 7 Ã— 3^{2},â€¦â€¦..**Solution:**

Let T

_{n}donotes the nth term of the given series. Then,

T_{n} = [nth term of 3, 5, 7,â€¦.]

[nth term of 1, 2, 3,â€¦.]^{2}

= [3 +(n â€“ 1)^{2}][1 + (n â€“ 1).1]^{2}

= (2n â€“ 1) (n)^{2}

= n^{2}(2n â€“ 1) = 2n^{3} + n^{2 }

Hence S_{n} = 2âˆ‘ n^{3} + âˆ‘ n^{2 }

= 2.

=

=

=

# Question-68

**Find the sum of the following series upto n terms**

**Solution:**

T

_{n}=

** = **

** = **

Let **= **â€¦â€¦(i)

â‡’ 1 = A(n + 1) + Bn â€¦â€¦(ii)

[On multiplying both sides by n(n+1)]

To find A : Putting n = 0 in (ii), we get,

1 = A(0 + 1) â‡’ A = 1

To find B : Putting n = 0 in (ii), we get ** **

1 = B(-1) â‡’ B = -1

Putting these values of A and B in (i), the partial fractions are

=

or T_{n} =

Putting n = 1, 2, 3,â€¦â€¦n, we get,

T_{1} =

T_{2} =

T_{3} = â€¦â€¦â€¦â€¦â€¦â€¦â€¦.

T_{n} =

Adding vertically, we get,

S_{n} = 1 - =

Extra. Hence find sum to infinity:

Now, S_{n }= =

As n â†’ âˆž

or Sâˆž = = 0

# Question-69

**Find the sum of the following series upto n terms**

**5**

^{2}+ 6^{2}+ 7^{2}+ â€¦â€¦+ 20^{2}**Solution:**

T

_{n}of given series

= (nth term of 5, 6, 7,â€¦..)^{2}

= [5 + (n + 1).1]^{2 }

_{= (n + 4)2 = n2 + 8n + 16 }

_{ }

_{Sn = âˆ‘ n2 + 8âˆ‘ n + 16 Ã— n }

_{ }

_{ }

_{ = }

_{ }

_{ = }

_{ }

_{ = }

_{ }

_{ = }

_{ }

_{Put n = 16, then}

_{ }

_{S = }

_{ }

_{ = }

_{ }

_{ = }

_{ }

_{ = 8 Ã— 355 = 2840. }

# Question-70

**Find the sum of the following series upto n terms**

**3 Ã— 8 + 6 Ã— 11 + 9 Ã— 14+â€¦.**

**Solution:**

Here the series is formed by multiplying the corresponding terms of two series both of which are A.P.

Viz. 3, 6, 9 â€¦â€¦ and 8, 11, 14â€¦..

T_{n} of given series

= (nth term of 3, 6, 9,â€¦.) Ã— (nth term of 8, 11, 14,â€¦.)

= [3 +(n â€“ 1)3] [8 + (n â€“ 1)3]

= [3n][3n + 5]

= 9n^{2} + 15n

S_{n} = 9 âˆ‘ _{n}^{2} + 15âˆ‘ _{n}

= 9 Ã—

=

= = 3n(n + 1) ( n + 3)

# Question-71

**1**

^{2}+ (1^{2}+ 2^{2}) + (1^{2}+ 2^{2}+ 3^{2}) + â€¦â€¦**Solution:**

a

_{n}= 1

^{2}+ 2

^{2}+ â€¦ + n

^{2}= = =

S_{n} =

=

=

= (n(n+1) + 2n + 1+ 1)

= (n^{2}+n+2n+1+1)

= (n^{2}+3n+2)

= (n+1)(n+2)

=

# Question-72

**Find the sum to Î¼ terms of the series in Exercise 8 to 10 whose nth terms is given by n(n+1)(n + 4)**

**Solution:**

T

_{n}= n(n+1)(n + 4)

T_{n} = n(n^{2}+5n+4) = n^{3} + 5n^{2} + 4n

S_{n} = âˆ‘ _{n}^{3} + 5âˆ‘ n^{2} + 4âˆ‘ _{n}

=

=

=

=

=

=

# Question-73

**n**

^{2}+ 2^{n }**Solution:**

a

_{n}= n

^{2}+ 2

^{n}

S

_{n}=

=

=

=

=

= + 2(2^{n} - 1).

# Question-74

**Find the sum to n terms of the following series whose nth term is given by**

**(2n â€“ 1)**

^{2}.**Solution:**

T

_{n}= (2n â€“ 1)

^{2 }

T_{n} = 4n^{2} â€“ 4n + l

S_{n} = 4âˆ‘ n^{2} - 4âˆ‘ n + n

=

=

=

=

=

=