**Q1. Obtain the sum of all integers which are odd between 1 and 4001 including 1 and 4001.**

**Answer:**

The integers which are odd between 1 and 2001 are 1, 3, 5, 7, 9, ……………. , 3997, 3999, 4001.

The sequence is in A.P form.

a = 1 [1^{st} term]

Difference, d = 2

The standard equation of A.P,

a + (s – 1) d = 4001

1 + (s – 1) 2 = 4001

s = 2001

S_{S} =

Hence, the sum of all integers which are odd between 1 and 4001 is 4004001.

**Q2. Obtain the sum of all natural numbers which are lying between 10 and 100, and are multiples of 5.**

**Answer:**

All natural numbers which are lying between 10 and 100, and are multiples of 5, are 15, 20, 25, … , 95.

The sequence is in A.P form.

a = 15 [1^{st} term]

Difference, d = 5

The standard equation of A.P,

a + (s – 1) d = 95

15 + (s – 1) 5 = 95

s = 17

S_{S} =

Hence, the sum of all natural numbers which are lying between 10 and 100, and are multiples of 5 is 935.

**Q3. Prove that the 21 ^{st} term in the sequence is – 118 provided that the sequence is in A.P form, the sum of the starting 5 terms is one – fourth of the next 5 terms and 2 is the first term.**

**Answer:**

Given:

2 is the first term, a = 2

The sequence is in A.P form,

In A. P = 2, 2 + d, 2 + 2d, 2 + 3d ……

In A.P form, the sum of the starting 5 terms is one – fourth of the next 5 terms,

So, according to the given condition,

10 + 10d = (1 / 4) (10 + 35d)

40 + 40d = 10 + 35d

d = – 6

a_{21} = a + (21 – 1) d = 2 + 20 (- 6) = – 118

Hence proved

**Q4. Obtain the number of terms in A.P which are needed to get – 25 from the sum of – 6, (- 11 / 2), – 5, ….**

**Answer**:

Suppose,

In, A.P the sum of s terms = – 25

a = – 6 [1^{st} term]

Difference, d = (- 11 / 2) + 6 = (1 / 2)

**Q5. If m ^{th} term is 1 / n and n^{th }term is 1 / m provided that the sequence is in A.P form, prove that the sum of the first mn terms is (1 / 2) (mn + 1), where m **

**n.**

**Answer:**

Given,

m ^{th} term is 1 / n and n^{th }term is 1 / m

So, acc. to the above condition

m ^{th} term in A.P form = a_{m} = a + (m – 1) d = 1 / m ……. (1)

n ^{th} term in A.P form = a_{n} = a + (n – 1) d = 1 / n ……. (2)

(2) – (1), we get,

(n – 1) d – (m – 1) d =

(n – m) d =

d =

Substituting the value of d in equation (1), we get,

a + (m – 1)

a =

Hence proved

**Q6. Obtain the last term if the addition of some numbers in A.P. 25, 22, 19, is 116.**

**Answer:**

Addition of s terms in A.P be 116

Here, first term a = 25, d = – 3

n = 8 is considered

a_{8 }(last term) = a + (s – 1) d = 25 + 7 (- 3) = 25 – 21 = 4

Hence, the last term is 4

**Q7. In A.P, obtain the total to s terms whose n ^{th} term is 5n + 1.**

**Answer:**

Given,

In A.P, the total to s terms whose n^{th} term is 5n + 1.

n^{th} term = a_{n} = a + (n – 1) d

a_{n} = a + (n – 1) d = 5n + 1

a + nd – d = 5n + 1

By comparing the coefficient of n, we get

d = 5 and a – d = 1

a – 5 = 1

a = 6

**Q8. Obtain the common difference of a sequence in A.P, when the total of s terms is (ms + ns ^{2}), where m and n are constants.**

**Answer:**

Given,

As mentioned in the given condition,

^{2})

Considering the coefficients of the of s^{2} , we get,

d = 2n

Hence, in A.P the difference d = 2n

**Q9. The ratio of the total of s terms of two arithmetic progressions are 5s + 4 : 9 + 6 . Obtain the ratio of 18 ^{th} term.**

**Answer:**

Suppose the first terms are a_{1 }and a_{2} respectively, and the common differences be d_{1 }and d_{2} of the first two consecutive arithmetic progressions respectively.

As mentioned in the given condition,

18^{th} term =

Hence, 179 : 321 is the required ratio of the 18^{th} term.

**Q10. In an A.P, the total of starting m terms is equal to the total of starting n terms. Obtain the total of starting (m + n) terms.**

**Answer:**

m ^{th }term = S_{m} =

n ^{th }term = S_{n} =

As mentioned in the given condition,

Hence, the total of starting (m + n) terms is 0.

**Q11. . In an A.P, the total of starting m, n and o terms are p, q and r. Show that the pm(n–o)+qn(m–o)+ro(m–n)=0.**

**Answer:**

Given,

In an A.P, the total of starting m, n and o terms are p, q and r.

As mentioned in the given condition,

Now, equating the values of d in equation (4) and (5)

Hence proved

**Q12. The ratio of the total of r and s terms of two arithmetic progressions is r ^{2} : s^{2}. Show that the ratio of r ^{th} and s ^{th} term is (2r – 1) : (2s – 1).**

**Answer:**

Suppose the first term is a, and the common difference be d of the arithmetic progressions respectively.

As mentioned in the given condition,

Hence proved

**Q13. In an A.P, the total of n terms is 3p ^{2} + 5p and its r^{th} term is 164. Obtain the value of r.**

**Answer:**

Given,

S_{r} =

Suppose the first term is a, and the common difference be d of the arithmetic progressions respectively.

As mentioned in the given condition,

p ^{th }term = S_{p} = ^{2} + 5p

As mentioned in the given condition,

From equation (i), we get

8 + (r – 1) 6 = 164

(r – 1) 6 = 164 – 8

(r – 1) 6 = 156

(r – 1) = 26

r = 7

**Q14. Place five numbers between 8 and 26 in a way such that the sequence results in an A.P form.**

**Answer:**

Suppose Q_{1}, Q_{2}, Q_{ 3}, Q_{ 4 }and Q_{ 5} be the five required numbers.

First term, a = 8,

Last term, p = 26

s = 7,

p = a + (s – 1) d

26 = 8 + (7 – 1) d

6 d = 18

d = 3

Q_{1} = a + d = 8 + 3 = 11

Q_{2} = a + 2 d = 8 + 2.3 = 14

Q_{ 3 }= a + 3 d = 8 + 3.3 = 17

Q_{ 4 }= a + 4 d = 8 + 3.4 = 20

Q_{ 5} = a + 5 d = 8 + 3. 5 = 23

Hence, Q_{1}, Q_{2}, Q_{ 3}, Q_{ 4 }and Q_{ 5 }is equal to 11, 14, 17, 20 and 23 are the required numbers in an A.P

**Q15. Suppose in an A.M pm+qmpm–1+qm–1 lies between p and q terms, then obtain the value of m.**

**Answer:**

A.M of p and q =

As mentioned in the given condition,

**Q16. Insert n numbers between 1 and 31 in a way such that the sequence results in an A.P. The ratio is 5 : 9 of the 7 ^{th} and the (n – 1)^{th }term. Find the value of n.**

**Answer:**

Suppose Q_{1}, Q_{2}, Q_{ 3}, Q_{ 4.}….. _{ }Q_{ m}, be the required n numbers.

First term, a = 1,

Last term, p = 31

s = n + 2,

p = a + (s – 1) d

31 = 1 + (n + 2 – 1) d

30 = (n + 1) d

d =

Q_{1} = a + d

Q_{2} = a + 2 d

Q_{ 3 }= a + 3 d

Q_{ 4 }= a + 4 d …….

Q_{ 7} = a + 7 d

Q_{ n – 1} = a + (n – 1) d

As mentioned in the given condition,

Hence, the value of n = 14

**Q17. A boy will be repaying his loans and his first installment is Rs 100. What is the amount should he pay in the 30 ^{th} installment if he increases the installment by Rs 5 every month?**

**Answer:**

Given,

His first installment is Rs 100

His 2^{nd} installment is Rs 105 and third installment is Rs 110 and so on

The sequence of money paid by the boy is in an A.P form every month is

100, 105, 110, 115 and so on …….

a = 100 [First term]

Common difference, d = 5

a _{30 }= a + (30 – 1) d

= 100 + 29 (5)

= 245

Hence, the amount should be paid by him in the 30^{th} installment is Rs 245.

**Q18. In a polygon, 120 ^{o} is the smallest angle and 5^{o} is the difference between consecutive angles on the interior side. Obtain the number of sides the polygon has.**

**Answer:**

Given,

120^{o} is the smallest angle i.e., first term, a = 120

And 5^{o} is the difference, i.e., d = 5

Total of all angles of a polygon having s sides = 180^{o} (s – 2)