Contents

**Andhra Pradesh SSC Class 10 Solutions For Maths – Progressions (English Medium) **

These Solutions are part of AP SSC Class 10 Solutions for Maths. In this article, we have provided Andhra Pradesh SSC Class 10 Solutions For Maths Chapter 6 Progressions

### Exercise 6.1:

**Question 1:
**In which of tine following situations, does the list of numbers involved make an arithmetic progression, and why?

- The taxi fare after each km when the fare is 20 for the first km and rises by 8 for each additional km.
- The amount of air present in a cylinder when a vacuum pump removes 1/4 of tine air remaining in the cylinder at a time.
- The cost of digging a well, after every metre of digging, when it costs 150 for the first metre and rises by’ 50 for each subsequent metre.
- The amount of money in the account every year, when ‘10000 is deposited at compound interest at 8 % per annum.

**Solution :**

- Arithmetic progression. Since rise in the taxi fair for each additional kilometer is same.
- Not Arithmetic progression. Since the amount of air removed using vacuum pump is not same always, it’s in the ratio of the air remaining in the cylinder.
- Arithmetic progression. Since rise in the cost of digging the well fair for each subsequent metre is same.
- Not Arithmetic progression. The amount of money in the account every year is increasing by applying compound interest on the deposit. In compound interest the interest gets added to the principle deposit for the next cycle, so every year the amount of money in the account is not same.

AP SSC 10th Class Textbook Solutions

**Question 2:
**Write first four terms of the AP, when the first term a and the common difference d are given as follows:

- a =10, d= 10
- a = -2, d=0
- a = 4, d = – 3
- a = – 1, d = 1/2
- a = – 1.25, d = – 0.25

**Solution :**

- The nth term of an AP having first term a and common difference d is given by

a_{n }= a + (n – 1)d

a_{1 }= 10 + (1 – 1)10 = 10

a_{2 }= 10 + (2 – 1)10 = 20

a_{3 }= 10 + (3 – 1)10 = 30

a_{4 }= 10 + (4 – 1)10 = 40 - The nth term of an AP having first term a and common difference d is given by

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

a_{1 }= -2 + (1 – 1)0 = -2

a_{2 }= -2 + (2 – 1)0 = -2

a_{3 }= -2 + (3 – 1)0 = -2

a_{4 }= -2 + (4 – 1)0 = -2 - The nth term of an AP having first term a and common difference d is given by

a_{n }= a + (n – 1)d

a_{1 }= 4 + (1 – 1)(-3) = 4

a_{2 }= 4 + (2 – 1)(-3) = 1

a_{3 }= 4 + (3 – 1)(-3) = -2

a_{4 }= 4 + (4 – 1)( -3) = -5 - The nth term of an AP having first term a and common difference d is given by

a_{n }= a + (n – 1)d

- The nth term of an AP having first term a and common difference d is given by

a_{n }= a + (n – 1)d

a_{1 }= -1.25 + (1 – 1)(-0.25) = -1.25

a_{2 }= -1 + (2 – 1)(-0.25) = -1.5

a_{3 }= -1 + (3 – 1)(-0.25) = -1.75

a_{4 }= -1 + (4 – 1)(-0.25) = -2

**Question 3:
**For the following APs, write the first term and the common difference:

- 3, 1,-1,-3,…
- – 5, – 1, 3, 7,…
- \(\frac { 1 }{ 3 } ,\frac { 5 }{ 3 } ,\frac { 9 }{ 3 } ,\frac { 13 }{ 3 },\)…….
- 0.6, 1.7, 2.8, 3.9,………..

**Solution :
**For an AP having the terms a

_{1}, a

_{2}, a

_{3},…

The first term a is given by,

a = a

_{1 }The common difference d is given by,

d = a

_{k+1 }– a

_{k}

**Question 4:
**Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

**Solution :**

- We have,

a_{2 }– a_{1 }= 4 – 2 = 2

a_{3 }– a_{2}= 8 – 4 = 4

As a_{2 }– a_{1 }¹ a_{3 }– a_{2}, the given list of numbers do not form an AP. - We have,

- We have,

a_{2 }– a_{1 }= -3.2 + 1.2 = -2

a_{3 }– a_{2 }= -5.2 + 3.2 = -2

i.e., the term a_{k+1 }– a_{k}remains same every time.

So, the given list of numbers forms an AP with the common difference d = -2.

The next three terms are -9.2, -11.2, -13.2. - We have,

a_{2 }– a_{1 }= -6 + 10 = 4

a_{3 }– a_{2 }= -2 + 6 = 4

i.e., the term a_{k+1}– a_{k}remains same every time.

So, the given list of numbers forms an AP with the common difference d = 4.

The next three terms are 6, 10, 14. - We have,

- We have,

a_{2 }– a_{1 }= 0.22 – 0.2 = 0.02

a_{3 }– a_{2 }= 0.222 – 0.22 = 0.002

As a_{2 }– a_{1 }¹ a_{3 }– a_{2}, the given list of numbers do not form an AP. - We have,

- We have,

- We have a
_{2 }– a_{1 }= 3 – 1 = 2

a_{3 }– a_{2 }= 9 – 3 = 6

As a_{2 }– a_{1 }¹ a_{3 }– a_{2}, the given list of numbers do not form an AP. - We have a
_{2 }– a_{1 }= 2a – a = a

a_{3 }– a_{2 }= 3a – 2a = a

i.e., the term a_{k+1 }– a_{k}remains same every time.

So, the given list of numbers forms an AP with the common difference d = a.

The next three terms are 5a, 6a, 7a. - We have a
_{2}– a_{1 }= a^{2 }– a

a_{3 }– a_{2 }= a^{3 }– a^{2}

As a_{2 }– a_{1 }¹ a_{3 }– a_{2}, the given list of numbers do not form an AP. - We have,

- We have,

### Exercise 6.2:

**Question 1:
**Fill in the blanks in the following table, given that a is the first term, d the common difference and a

_{n}the nth term of the AP:

**Solution :**

Therefore, the number of terms n of the given A.P. is 10.

v. Here , a = 3.5, d = 0 and n = 105

We have a

_{n }= a + (n – 1)d

So, a

_{105 }= 3.5 + (105 – 1)(0) = 3.5 + 0 = 3.5.

Therefore, the n

^{th}term a

_{n}of the given A.P. is 3.5.

Sr. No |
a |
d |
n |
a_{n} |

1 | 7 | 3 | 8 | 28 |

2 | -18 | 2 | 10 | 0 |

3 | 46 | -3 | 18 | -5 |

4 | -18.9 | 2.5 | 10 | 3.6 |

5 | 3.5 | 0 | 105 | 3.5 |

** **

**Question 2:
**Find the

- 30
^{th}term of the A.P. 10, 7, 4……. - 11
^{th}term of the A.P. : \(-3,\frac { -1 }{ 2 } ,2,\)……..

**Solution :
**i. Here , a = 10, d = 7 – 10 = -3 and n = 30

We have a

_{n }= a + (n – 1)d

So, a

_{30 }= 10 + (30 – 1)(-3) = 10 – 87 = -77.

Therefore, the 30

^{th}term of the given A.P. is -77.

**Question 3:
**Find the respective terms for the following APs.

**Solution :**

- Since the terms are in A.P., so,

a_{3 }– a_{2 }= a_{2 }– a_{1}

a_{3 }– a_{2 }– a_{2 }+ a_{1 }= 0

a_{3 }– 2a_{2 }+ a_{1 }= 0

a_{3 }+ a_{1 }= 2a_{2}

26 + 2 = 2a_{2}

28 = 2a_{2}

a_{2 }= 14 - Since the terms are in A.P., so,

a_{4 }– a_{3 }= a_{3 }– a_{2}

a_{4 }– a_{3 }– a_{3 }+ a_{2 }= 0

a_{4 }– 2a_{3 }+ a_{2 }= 0

a_{4 }+ a_{2 }= 2a_{3}

13 + 3 = 2a_{3}

16 = 2a_{3}

a_{3 }= 8

so, a_{2 }– a_{1 }= a_{3 }– a_{2}

_{ }a_{2 }– a_{1 }– a_{3 }+ a_{2 }= 0

2a_{2 }– a_{3 }= a_{1}

_{ }26 – 8 = a_{1}

a_{1 }= 18 - a
_{1 }= a = 5

- a
_{1 }= a = -4

a_{6 }= a + (n-1)d

6 = a + (6-1)d

6 = a + 5d

6 = -4 + 5d

10 = 5d

d =2

a_{2 }= a + (n-1)d = -4+(2-1)2=-4+2=-2

a_{3}=a+(n-1)d=-4+(3-1)2=-4+4=0

a_{4}=a+(n-1)d=-4+(4-1)2=-4+6=2

a_{5}=a+(n-1)d=-4+(5-1)2=-4+8=4 - a
_{2}=a_{1}+(n-1)d

38= a_{1}+d

a_{6}=a_{1}+(n-1)d

-22= a_{1}+5d

a_{6}– a_{2}=4d

-22-38=4d

-60=4d

d=-15

a_{2}=a_{1}+d

a_{1}=a_{2}-d

a_{1}=38+15

a_{1}=53=a

a_{3}=a+(n-1)d=53+(3-1) (-15) = 53-30=23

a_{4}=a+(n-1)d=53+(4-1) (-15) = 53-45=8

a_{5}=a+(n-1)d=53+(5-1) (-15) = 53-60=-7

**Question 4:
**Which term of the AP : 3, 8, 13, 18,……,is 78?

**Solution :**

Here, a=3, d=8-3=5, and if a

_{n}=78, we have to find n.

As, a

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

78=3+(n-1)5

78=3+5n-5

83=3+5n

80=5n

n = 16

Therefore, the 16

^{th}term of the given A.P. is 78.

**Question 5:
**Find the number of terms in each of the following APs :

(i) 7, 13, 19,… ,205

(ii) \(18,\quad 15\frac { 1 }{ 2 } ,13,\)……..,-47

**Solution :**

i. Here a=7, d=13-7=6 , and if a

_{n}=205

As a

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

205=7+(n-1)6

205=7+6n-6

205=1+6n

204=6n

n=34

**Question 6:
**Check whether, -150 is a term of the AP: 11,8,5,2…

**Solution :**

We have:

a

_{2}-a

_{1}=8-11=-3, a

_{3}-a

_{2}=5-8=-3

As (a

_{k+1}-a

_{k}) is the same for k=1, 2, etc, the given list of numbers is an A.P.

Now, for this A.P. we have a = 11 and d = -3.

Let the n

^{th}term of this AP be -150.

We know,

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

-150=11+(n-1)(-3)

-150=11-3n+3

-150=14-3n

-164=-3n

n=164 ⁄ 3

But n should be a positive integer.

Hence, -150 is not a term of the given list of numbers.

**Question 7:
**Find the 31

^{st}term of an AP whose 11

^{th }term is 38 and the 16

^{th }term is 73.

**Solution :**

We have,

a

_{11}=a+(11-1)d=a+10d=38 (1)

a

_{16}=a+(16-1)d=a+15d=73 (2)

Solving equation (1) and (2), we get

a= -32 and d=7

Hence, the 31

^{st}term will be,

a

_{31}=a+(n-1)d

=-32+(31-1)7

=-32+(30)7

=-32+210

=178

Therefore the 31

^{st}term of the A.P. is 178.

**Question 8:
**If the 3

^{rd}and the 9

^{th}terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

**Solution :**

a

_{3}=a+(n-1)d

4=a+(3-1)d

4=a+2d (1)

a

_{9}=a+(n-1)d

-8=a+(9-1)d

-8=a+8d (2)

Solving (1) and (2), we get

a = 8 and d=-2

We want to know if there is any n for which a

_{n}=0.

a

_{n}=

_{}a+(n-1)d

0=8+(n-1)(-2)

0=10-2n

n=5

**Question 9:
**The 17

^{th}term of an AP exceeds its 10

^{th}term by 7. Find the common difference.

**Solution :**

Let the first term of the A.P. be ‘a’, 17

^{th}term be ‘a

_{17}‘,

10

^{th}term be ‘a

_{10}‘, and common difference be ‘d’.

a

_{17}=a+16d—–(1)

a

_{10}=a+9d——-(2)

Subtracting (2) from (1)

a

_{17}-a

_{10}=a+16d-(a+9d)

7=a+16d-a-9d

7=7d

d=1

Thus, the common difference of the given A.P. is 1.

**Question 10:
**Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

**Solution :**

Let the first term of the AP1 be ‘a1’, 100

^{th}term be ‘a1

_{100}‘, and 1000

^{th}term be ‘a1

_{1000}‘.

Let the first term of the AP2 be ‘a2’, 100

^{th}term be ‘a2

_{100}‘, 1000

^{th}term be ‘a2

_{1000}‘,

Let the common difference be ‘d’, as it is same for both the APs.

a1

_{100}=a1+99d—–(1)

a2

_{100}=a2+99d—–(2)

Subtract (2) from (1)

a1

_{100}-a2

_{100}=a1+99d-a2-99d

a1

_{100}-a2

_{100}=a1-a2—–(3)

But,

a1

_{100}-a2

_{100}=100 ……..( Given)

Therefore,

a1-a2=100——(3)

Now,

a1

_{1000}=a1+999d—–(4)

a2

_{1000}=a2+999d—–(5)

Subtract (5) from (4)

a1

_{1000}-a2

_{1000}=a1+999d-a2-999d

a1

_{1000}-a2

_{1000}=a1-a2

But ,

a1-a2=100 ………..( from (3))

Therefore,

a1

_{1000}-a2

_{1000}=100

Thus the difference between the 1000

^{th}term of both the APs is also 100, as proved above.

**Question 11:
**Flow many three-digit numbers are divisible by 7?

**Solution :**

Three digit numbers which are divisible by 7 are:

105, 112, ………..994

So, a=105, d=112-115=7, a

_{n}=994

a

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

994=105+(n-1)7

994=105+7n-7

994-98=7n

896=7n

n=128

There are 128 three digit numbers divisible by 7.

**Question 12:
**Flow many multiples of 4 lie between 10 and 250?

**Solution :**

The multiples of 4 which lie between 10 and 250 are,

12, 16, 20, 24, 28……….248

So, a=12, d=16-12=4 and a

_{n}=248

a

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

a

_{n}=12+(n-1)4

248 =12+4n-4

248=8+4n

240=4n

n=6.

**Question 13:
**For what value of n, are the nth terms of two APs: 63, 65, 67,… and 3, 10, 17,… equal?

**Solution :**

For the first A.P., 63, 65, 67….

The first term a

_{1}=63 and d

_{a}=65-63=2

a

_{n}= a

_{1}+(n-1) d

_{a }For the second A.P., 3, 10, 17….

The first term b

_{1}=3 and d

_{b}=10-3=7

b

_{n}= b

_{1}+(n-1) d

_{b }Let the n

^{th}term of A.P1 and A.P2 be equal.

a

_{n}= b

_{n }a

_{1}+(n-1) d

_{a}= b

_{1}+(n-1) d

_{b }63+(n-1)2=3+(n-1)7

60=(n-1)5

n-1=12

n=13

So, the 13

^{th}term of A.P1 and A.P2 are equal.

**Question 14:
**Determine the AP whose third term is 16 and the 7

^{th}term exceeds the 5

^{th}term by 12.

**Solution :**

Let the first term of the A.P. be ‘a’ and the common difference be ‘d’.

The third term of the A.P. is 16.

a

_{3}= a+(3-1) d

16=a+2d…..(i)

The 7

^{th}term exceeds the 5

^{th}term by 12.

a

_{7}-a

_{5}=12

a+(7-1)d-[ a+(5-1)d]=12

a+6d-a-4d=12

2d=12

d=6

Substitute d=6 in equation (i)

16=a+2(6)

a=4

a

_{1}=a=4

a

_{2}=a+(n-1)d=4+(2-1) (6) =4+6=10

a

_{3}= 16

Hence the A.P. is 4, 10, 16,….

**Question 15:
**Find the 20

^{th}term from the end of the AP : 3, 8,13,…, 253.

**Solution :**

Let the first term of the A.P. be ‘a’ and the common difference be ‘d’.

a=3, d=8-3=5 and a

_{n}=253

The n

^{th}term of the A.P. is given by,

a

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

253= 3+(n-1)5

250=(n-1)5

n-1=50

n=51

The 20

^{th}term from the end of the A.P. is the 32

^{th}term from the top of the A.P., given by,

a

_{31}= a+(32-1) d

a

_{31}= 3+(32-1)5

a

_{31}=3+155

a

_{31}= 158

**Question 16:
**The sum of the 4 and 8 terms of an AP is 24 and the sum of the 6 and 10 terms is 44. Find the first three terms of the AP.

**Solution :**

Let the first term of the A.P. be ‘a’ and the common difference be ‘d’.

The sum of the 4

^{th}and 8

^{th}terms of the A.P. is 24,

a

_{4 }+ a

_{8}=24

a+(4-1)d+a+(8-1)d =24

2a+10d=24

a+5d=12…………(i)

The sum of the 6

^{th}and 10

^{th}term of the A.P. is 44,

a

_{6 }+ a

_{10}=44

a+(6-1)d+a+(10-1)d =44

2a+14d=44

a+7d=22…………(ii)

Subtracting equation (i) from (ii) we get

2d=10

d=5

Substituting d=5 in equation (i) we get

a+5(5)=12

a=-13

The first three terms of the A.P. are given by,

a

_{1}= a=-13

a

_{2}= -13+(2-1)5=-8

a

_{3}=-13+(3-1)5=-3

**Question 17:
**Subba Rao started work in 1995 at an annual salary’ of’ 5000 and received an increment of’ 200 each year. In which year did his salary reach 7000?

**Solution :**

Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year.

So the annual salary from 1995 onwards is

5000, 5200, 5400,……

Let the first term of the A.P. be ‘a

_{1}‘ and the common difference be ‘d’.

a

_{1 }= 5000, d = 200

To find the value of n for which a

_{n }= 7000

The n

^{th}term of the A.P. is given by

a

_{n}= a

_{1}+(n-1) d

7000= 5000+(n-1) 200

2000= (n-1) 200

n-1=10

n=11

In the year 2005, Subba Rao’s income reaches Rs. 7000.

### Exercise 6.3:

**Question 1:
**Find the sum of the following APs:

(i) 2, 7, 12,..to 10 terms.

(ii) -37,-33,-29,…. to 12 terms

(iii) 6, 1.7, 2.8,…, to 100 terms.

(iv) \(\frac { 1 }{ 15 } ,\frac { 1 }{ 12 } ,\frac { 1 }{ 10 } ,\)….,to 11 terms

**Solution :**

**Question 2:
**Find the sums given below:

**Solution :**

**Question 3:
**In an AP:

**Question 3(i):
**given a = 5, d=3, a =50, find n and S

_{n}.

**Solution :**

**Question 3(ii):
**given a = 7, a

_{13}= 35, find d and S

_{13. }

**Solution :**

**Question 3(iii):
**given a

_{12}= 37, d= 3, find a and S

_{12}.

**Solution :**

**Question 3(iv):
**given a

_{3}= 15, S

_{10}= 125, find d and a

_{10}.

**Solution :**

**Question 3(v):
**given a = 2, d = 8, S

_{n}= 90, find n and a

_{n}.

**Solution :**

**Question 3(vi):
**given a

_{n}= 4, d = 2, S

_{n}= -14, find n and

**Solution :**

**Question 3(vii):
**given

*l*= 28, S = 144, and there are total 9 terms. Find a

**Solution :**

**Question 4:
**The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

**Solution :**

**Question 5:
**Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

**Solution :**

**Question 6:
**If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n

**Solution :**

**Question 7:
**Show that a

_{1,}a

_{2},……..,a

_{n},… form an AP where a

_{n}is defined as below :

(i) a

_{n}= 3 + 4n

(ii) a

_{n}= 9 – 5n

Also find the sum of the first 15 terms in each case.

**Solution :**

**Question 8:
**If the sum of the first n terms of an AP is 4n – n

^{2}, what is the first term (remember the first term is S

_{1})? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

**Solution :**

**Question 9:
**Find the sum of the first 40 positive integers divisible by 6.

**Solution :**

**Question 10:
**A sum of’ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is’ 20 less than its preceding prize, find the value of each of the prizes.

**Solution :**

**Question 11:
**In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

**Solution :**

Initially, just considering one section per class,

if each class plants the same number of trees as the level/standard/grade of it, then the trees planted by successive classes can be considered to be an A.P. with a = 1(since class starts with Std I) and d = 1.

Since there are 12 classes,

Therefore,

a

_{12 }= a + 11d

a

_{12 }= 1 + 11(1)

a

_{12 }= 12———-(1)

Sum of trees planted by one section of each class is given as,

S

_{12 }= 6 (a + a

_{12})

S

_{12 }= (6)(1 + 12)

S

_{12 }= 78———-(2)

Therefore, when 3 sections per class are considered, then the total number of trees are,

3xS

_{12}=3×78=234

Hence, the total numbers of trees planted by the children is 234.

**Question 12:
**A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen

consecutive semicircles? (Take π= 22/7 )

[Hint: Length of successive semicircles is l

_{1,}l

_{2},l

_{3,}l

_{4,}…… with centres at A, B, A, B,…..respectively.]

**Solution :**

**Question 13:
**200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

**Solution :**

**Question 14:
**In a bucket and ball race, a bucket is placed at the starting point, which is 5 m from the first ball, and the other balls are placed 3 m apart in a straight line. There are ten balls in the line.

A competitor starts from the bucket, picks up the nearest ball, runs back with it, drops it in the bucket, runs back to pick up the next ball, runs to the bucket to drop it in, and she continues in the same way until all the balls are in the bucket. What is the total distance the competitor has to run?

[Hint: To pick up the first ball and the second ball, the total distance (in metres) run by a competitor is

2 x 5 + 2 x (5 + 3)]

**Solution :**

### Exercise 6.4:

**Question 1:
**I Which of the following situations, does the list of numbers involved in the form of a GP.?

- Salary of Sharmila, when her salary is 5,00,000 for the first year and expected to receive yearly increase of 10%.
- Number of bricks needed to make each step, if the stair case has total 30 steps. Bottom step needs 100 bricks and each successive step needs 2 brick less than the previous step.
- Perimeter of the each triangle, when the mid points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid points in turn are joined to form still another triangle and the process continues indefinitely.

**Solution :
**

**Question 2:
**Write three terms of the GP when the first term ‘a’ and the common ratio ‘r’ are given?

(i) a = 4; r = 3

(ii) a = √5; r = 1/5

(iii) a = 81; r =-1/3

(iv) a= 1/64; r = 2

**Solution :**

**Question 3:
**Which of the following are GP? If they are in GP Write three more terms?

**Question 3(i):
**4, 8, 16…..

**Solution :**

We know that a list of numbers a_{1}, a_{2, }a_{3}……a_{n}… is called a GP if each term is non zero and

So, the given list of the number forms a GP with the first term as a=4 and the common ratio r=2.

Thus the next three terms are:

ar^{3} , ar^{4}, ar^{5} = 32, 64, 128

**Question 3(ii):
**\(\frac { 1 }{ 3 } ,\frac { -1 }{ 6 } \frac { 1 }{ 12 }\) ……

**Solution :**

We know that a list of numbers a

_{1}, a

_{2, }a

_{3}……a

_{n}… is called a GP if each term is non zero and

**Question 3(iii):
**5, 55, 555,………

**Solution :**

We know that a list of numbers a

_{1}, a

_{2, }a

_{3}……a

_{n}… is called a GP if each term is non zero and

So, the given list of the numbers does not form a GP

**Question 3(iv):
**-2, -6, -l8 …….

**Solution :**

We know that a list of numbers a

_{1}, a

_{2, }a

_{3}……a

_{n}… is called a GP if each term is non zero and

So, the given list of the numbers form a GP with the first term as A = -2 and common ratio, r = 3.

Thus the next three terms are:

ar^{3}, ar^{4}, ar^{5} = -54, -162, -486

**Remark:** The answer given in the text book is incorrect.

**Question 3(v):
**\(\frac { 1 }{ 2 } ,\frac { 1 }{ 4 } ,\frac { 1 }{ 6 }\) …….

**Solution :**

We know that a list of numbers a

_{1}, a

_{2, }a

_{3}……a

_{n}… is called a GP if each term is non zero and

So, the given list of the numbers does not form a GP

**Question 3(vi):
**\(3,-{ 3 }^{ 2 },{ 3 }^{ 3 },\)….

**Solution :**

We know that a list of numbers a

_{1}, a

_{2, }a

_{3}……a

_{n}… is called a GP if each term is non zero and

So, the given list of the number forms a GP with the first term as a=3 and common ratio r=-3.

The next three terms are :

Ar

^{3}, ar

^{4}, ar

^{5}= -81, 243, -729

**Question 3(vii):
**\(x,1,\frac { 1 }{ x } ,\)…..

**Solution :**

We know that a list of numbers a

_{1}, a

_{2, }a

_{3}……a

_{n}… is called a GP if each term is non zero and

**Question 3(viii):
**1/√2, -2, 8/√2……..

**Solution :**

We know that a list of numbers a

_{1}, a

_{2, }a

_{3}……a

_{n}… is called a GP if each term is non zero and

**Question 3(ix):
**0.4, 0.04, 0.004,…….

**Solution :**

We know that a list of numbers a

_{1}, a

_{2, }a

_{3}……a

_{n}… is called a GP if each term is non zero and

So, the given list of the number forms a GP with the first term as a=0.4 common ratio r= 0.1

Thus, the next three terms are:

ar

^{3}, ar

^{4}, ar

^{5}= 0.0004, 0.00004, 0.000004

**Question 4:
**Find x so that x, X + 2,x + 6 are consecutive terms of a geometric progression.

**Solution :**

Since they are consecutive terms and hence b=3

So , x, x+2, x+3

(x+2)

^{2}=x(x+3)

x

^{2}+4x+4=x

^{2}+3x

4x+4=3x

4x-3x=-4

x=-4

**Ans:**the value of x is -4.

### Exercise 6.5:

**Question 1:
**For each geometric progression find the common ratio ‘r’, and then find a

_{n. }(i) \(3,\frac { 3 }{ 2 } ,\frac { 3 }{ 4 } ,\frac { 3 }{ 8 } \)…….

(ii) 2,-6, 18,-54

(iii) -1,-3, -9,-27….

(iv) \(5,2,\frac { 4 }{ 5 } ,\frac { 8 }{ 25 } \)………

**Solution :**

**Question 2:
**Find the 10

^{th}and n

^{th}term of GP.: 5,25,125,……….

**Solution :**

Here a=5 and r=5

In a GP nth term is given by a

_{n}=ar

^{n-1}.

Then a

_{10}=5(5)

^{10-1}=5(5)

^{9}=5

^{10 }and a

_{n}=ar

^{n-1}=5(5)

^{n-1}=5

^{n}

**Question 3:
**Find the indicated term of each geometric Progression

**Solution :**

**Question 4:
**Which term of the GP.

**Solution :**

**Question 5:
**Find the 12

^{th}term of a GP. whose 8

^{th}term is 192 and the common ratio is 2.

**Solution :**

**Question 6:
**The 4th term of a geometric progression is 2/3 and the seventh term is 16/81. Find the geometric series.

**Solution :**

**Question 7:
**If the geometric progressions 162, 54,18 and \(\frac { 2 }{ 81 } ,\frac { 2 }{ 27 } ,\frac { 2 }{ 9 } \)…. have their n

^{th}term equal. Find the value of n.

**Solution :**

Hope given Andhra Pradesh SSC Class 10 Solutions For Maths chapter 6 Progressions are helpful to complete your math homework.

If you have any doubts, please comment below. CBSEtuts try to provide online math tutoring for you.