Exercise 5.1
Multiple Choice Questions (Q. No. 1-9)
Question. 1 In an AP, if d = − 4, n = 7 and a_n = 4, then a is equal to
(a) 6
(b) 7
(c) 20
(d) 28
Solution. (d)
Explanation:
We know that in an AP,
     a_n = a + (n − 1) d
⟹ 4 = a + (7 − 1) (− 4)                       [by given conditions]
⟹ 4 = a + 6 (− 4)
⟹ a = 4 + 24 = 28
Question. 2 In an AP, if a = 3.5, d = 0 and n = 101, then a_n will be
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5
Solution. (b)
Explanation:
We know that in an AP,
      a_n = a + (n − 1) d
⟹ a_n = 3.5 + (101 − 1) × 0     [by given conditions]
⟹ a_n = 3.5
Question. 3 The list of numbers − 10, − 6, − 2, 2, … is
(a) an AP with d = −16
(b) an AP with d = 4
(c) an AP with d = − 4
(d) not an AP
Solution. (b)
Explanation:
Given list of numbers is: −10, − 6, − 2, 2,….
Here, a₁ = −10, a₂ = − 6, a₃ = − 2 and a₄ = 2,
Now, d₁ = a₂ – a₁ = − 6 − (−10) = − 6 + 10 = 4
d₂ = a₃ – a₁ = − 2 − (− 6) = − 2 + 6 = 4
And d₃ =  a₄ – a₃ = 2 − (− 2) = 2 + 2 = 4
As d₁ = d₂ = d₃ = 4
So, the given list forms an AP with common difference, d = 4.

(a) –20
(b) 20
(c) –30
(d) 30
Solution. (b)
Explanation:

Question. 5 The first four terms of an AP whose first term is − 2 and the common difference is − 2 are
(a) − 2, 0, 2, 4
(b) − 2, 4, − 8, 16
(c) − 2, − 4, − 6, − 8
(d) − 2, − 4, − 8, − 16
Solution. (c)
Explanation:
Given, a₁ = –2 and  d = –2
Therefore, a₂ = a₁ + d
⟹       a₂ = –2 – 2 = – 4
Also     a₃ = a₂ + d = – 4 + (–2) = –6
And     a₄ = a₃ + d = – 6 + (–2) = –8
So, the first four terms are –2, –4, –6, –8.
Question. 6 The 21st term of an AP whose first two terms are − 3 and 4, is
(a) 17
(b)137
(c) 143
(d) −143
Solution. (b)
Explanation:
For an AP, a_n = a + (n – 1)d  …..(i)
Given, first two terms of an AP are − 3 and a + d = 4.
⟹ a = − 3 and a₁= −3 + d = 4
⟹ Common difference, d = a1 – a = 4 – (−3) = 4+3 = 7
Therefore, a₂₁ = a + (21 − 1) d            [Using (i)]
                       = − 3 + (20) 7
                       = − 3 + 140 = 137
Question. 7 If the 2nd term of an AP is 13 and 5th term is 25, what is its 7th term?
(a) 30
(b) 33
(c) 37
(d) 38
Solution. (b)
Explanation:
Given, a₂ = 13 and a₅ = 25
Using a_n = a + (n – 1)d, we have:
      a + (2 − 1) d = 13 and      a + (5 − 1) d = 25
⟹  a + d = 13 ….(i)  and a + 4 d = 25   …..(ii)
On subtracting equation (i) from equation (ii), we get
     3 d = 25 – 13 = 12
⟹  d = 4
Putting value of d in equation (i), we get:
       a = 13 − 4 = 9
⟹  a₇ = a + (7 − 1) d = 9 + 6 × 4 = 33
Question. 8 Which term of an AP : 21, 42, 63, 84, … is 210?
(a) 9^th
(b) 10^th
(c) 11^th
(d) 12^th
Solution. (b)
Explanation:
Let nth term of the given AP be 210.
Here, first term, a = 21
Common difference, d = 42 − 21 = 21
And a_n = 210
Using a_n = a + (n – 1)d, we have:
       210 = 21 + (n – 1) 21
⟹   210 = 21+ 21n – 21
⟹   210 = 21 n
⟹      n = 10
Hence, the 10th term of an AP is 210.
Question. 9 If the common difference of an AP is 5, then what is a₁₈ – a₁₃?
(a) 5
(b) 20
(c) 25
(d) 30
Solution. (c)
Explanation:
Given, common difference of an AP, d = 5
Using a_n = a + (n – 1)d, we have:
        a₁₈ =a + (18 − 1) d
And  a₁₃ = a + (13 − 1) d
Thus, a₁₈ – a₁₃ = [a + (18 – 1) d] – [a + (13 -1) d]                   
                        = a + 17 × 5 – a – 12 × 5
                        = 85 – 60 = 25
Exercise 5.1
Multiple Choice Questions (Q. No. 9-18)
Question. 10 What is the common difference of an AP in which a₁₈ − a₁₄ = 32?
(a) 8
(b) – 8
(c) – 4
(d) 4
Solution. (a)
Explanation:
Given, a₁₈ – a₁₄ = 32
Using a_n = a + (n – 1)d, we have:
            a₁₈ =a + (18 − 1) d
And a₁₃ = a + (13 − 1) d
Thus, a₁₈ – a₁₃ = [a + (18 − 1)d] − [a + (14 − 1) d] = 32
⟹       a + 17d – a – 13d = 32
⟹       4d = 32
⟹       d = 8
Thus, the required common difference of given AP = 8.
Question. 11 Two APs have the same common difference. The first term of one ofthese is − 1 and that of the other is − 8. The difference between their 4th terms is
(a) – 1
(b) – 8
(c) 7
(d) – 9
Solution. (c)
Explanation:
Let the same common difference of two APs be d.
Given, first term of first AP (a₁) = − 1
and the first term of second AP (a₁’) = − 8
We know that, the nth term of an AP is given as,
            a_n = a + (n − 1) d
Therefore, 4th term of first AP is,
            a₄ = a₁ + (4 − 1) d = −1 + 3d
And 4th term of second AP is,
            a₄’ = a₁’ + (4 − 1) d = − 8 + 3d
Difference between 4th terms of both APs is:
            a₄ − a₄’ = (−1 + 3d) − (− 8 + 3d)
                        = − 1 + 3d + 8 − 3d = 7
Hence, the required difference is 7.
Question. 12 If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be
(a) 7
(b) 11
(c) 18
(d) 0
Solution. (d)
Explanation:
We know that, the nth term of an AP is given as,
            a_n = a + (n − 1) d
Now, according to the question,
                                 7a₇ = 11a₁₁
⟹       7 [a + (7 − 1) d] = 11 [a + (11-1) d]                           
⟹                  7 (a +6d) = 11 (a + 10d)
⟹                  7a + 42d = 11a + 110d
⟹                   4a + 68d = 0
⟹                4 (a +17d) = 0
⟹                       a + 7d = 0                            [4 ≠ 0]
Or              a + (18 – 1) d = 0  
⟹18^th term of an AP, a₁₈ = 0
Question. 13 The 4th term from the end of an AP − 11, − 8, − 5,…, 49  is
(a) 37
(b) 40
(c) 43
(d) 58
Solution. (b)
Explanation:
Taking the AP in reverse order: 49,…, −5, −8, −11.
Here, a_l = 49   
Common difference, d = − 8 − (− 11)
                                       = − 8 + 11= 3
Now, we know that, the nth term of an AP is given as:
            a_n = a_l – (n − 1) d       
Therefore, fourth term of AP is,
            a₄ = 49 − (4 − 1) 3 = 49 − 9 = 40
Question. 14 The famous mathematician associated with finding the sum of the first 100 natural numbers is
(a) Pythagoras
(b) Newton
(c) Gauss
(d) Euclid
Sol. (c)
Explanation:
Gauss is the famous mathematician associated with finding the sum of the first 100 natural numbers i.e., 1 + 2 + 3 +…+ 100.
Question. 15 If the first term of an AP is − 5 and the common difference is 2, then the sum of the first 6 terms is
(a) 0
(b) 5
(c) 6
(d) 15
Solution. (a)
Explanation:

Question. 16 The sum of first 16 terms of the AP 10, 6, 2,… is
(a) −320
(b) 320
(c) −352
(d) −400
Solution. (a)
Explanation:

Question. 17 In an AP, if a = 1, a_n = 20 and S_n = 399, then n is equal to
(a) 19
(b) 21
(c) 38
(d) 42
Solution. (c)
Explanation:

Question. 18 The sum of first five multiples of 3 is
(a) 45
(b) 55
(c) 65
(d) 75
Solution. (a)
Explanation:
The first five multiples of 3 are 3, 6, 9, 12 and 15.
3, 6, 9, 12 and 15 form an AP with both its first term and common difference (6 – 3) being 3.
i.e., a = 3 and d = 3
Also, number of terms, n = 5

Exercise – 5.2
Very Short Answer Type Questions
Question. 1 Which of the following form of an AP? Justify your answer.

Solution.
A series of numbers will form an AP, if d₁ = d₂ = d₃…
(i) −1, −1, −1, −1, ….
Here, t₁ = −1, t₂ = −1, t₃ = −1 and t₄ = − 1
Now,   d₁ = t₂ – t₁ = − 1 + 1 = 0
            d₂ = t₃ − t₂ = − 1 + 1 = 0
            d₃ = t₄ − t₃ = − 1 + 1 = 0
As d₁ = d₂ = d₃, therefore given list of numbers forms an AP.
(ii) 0, 2, 0, 2, …
Here, t₁ = 0, t₂ = 2, t₃ = 0 and t₄ = 2
Now,   d₁ = t₂ −  t₁ = 2 − 0 = 2
            d₂ = t₃ − t₂ = 0 − 2 = − 2
            d₃ = t₄ −  t₃ = 2 − 0 = 2
As d₁ ≠ d₂, therefore given list of numbers does not form an AP.
(iii) 1, 1, 2, 2, 3, 3, …..
Here, t₁ = 1, t₂ = 1, t₃ = 2 and t₄ = 2
Now,   d₁ = t₂ −  t₁ = 1 − 1 = 0
            d₂ = t₃ − t₂ = 2 – 1 = 1
            d₃ = t₄ − t₃ = 2 − 2 = 0
As d₁ ≠ d₂, therefore given list of numbers does not form an AP.
 (iv) 11, 22, 33, …..
Here, t₁ = 11, t₂ = 22 and t₃ = 33
Now,   d₁ =  t₂ − t₁ = 22 − 11 = 11
            d₂ = t₃ − t₂ = 33 − 22 = 11
            d₃ = t₄ − t₃ = 33 − 22 = 11
As d₁ = d₂ = d₃, therefore given list of numbers forms an AP.

As d₁ ≠ d₂, therefore given list of numbers does not form an AP.
(vi) 2, 2², 2³, 2⁴,…. i.e., 2, 4, 8, 16, ……
Here,   t₁ = 2, t₂ = 4, t₃ = 8 and t₄ = 16
Now,   d₁ = t₂ − t₁ = 4 − 2 = 2
And     d₂ = t₃ − t₂ = 8 − 4 = 4
As d₁ ≠ d₂, therefore given list of numbers does not form an AP.

As d₁ = d₂ = d₃, therefore given list of numbers forms an AP.

Since, here a₂ – a₁ = a₃ – a₂ but a₃ – a₂ ≠ a₄ – a₃, i.e., d₂ ≠ d₃ therefore given series of numbers does not form an AP.
Question. 3 For the AP −3, −7, −11,….. can we find directly a₃₀ – a₂₀ without actually finding a₃₀ and a₂₀? Give reason for your answer.
Solution.
Yes.
We know that nth term of an AP ia given as,
                        a_n = a + (n − 1) d
Therefore,        a₃₀ = a + (30 − 1) d = a + 29d
And                 a₂₀ = a + (20 – 1) d = a + 19d                                     
⟹                   a₃₀ – a₂₀ = (a + 29d) – (a + 19d) = 10d
Now, form given AP common difference, d = −7 – (−3) = −7 + 3 = −4
Therefore,        a₃₀ – a₂₀ = 10 (–4) = –40                                             
Question. 4 Two AP’s have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10^th terms is the same as the difference between their 21^st terms, which is the same as the difference any two corresponding terms. Why?
Solution.
Let the same common difference of two APs be d.
Given, first term of first AP, a₁ = 2
And, first term of second AP, a₁’= 7
Using, a_n = a + (n − 1)d,   
10^th term of first AP, a₁₀ = 2 + 9d
And 10^th term of second AP, a₁₀’ =7 + 9d
So, a₁₀ − a₁₀’ = (2 + 9d) −  (7 + 9d) = −5
Again using a_n = a + (n − 1)d,
21^st term of first AP, a₂₁ = 2 + 20d 
And 21^st term of second AP, a₂₁’ =  7 + 20d
So, a₂₁ − a₂₁’ = (2 + 9d) − (7 + 20d) = −5
Also, if the a_n and a_n’ are the nth terms of first and second AP.
Then, a_n – a_n’ = [2 + (n – 1)d] – [7 + (n – 1)d] = –5
Hence, the difference between any two corresponding terms of such AP’s is the same as the difference between their10^th and 21^st terms.
Question. 5 Is 0 a term of the AP 31, 28, 25,…? Justify your answer.
Solution.
Let 0 be the nth term of given AP, i.e., a_n = 0.
Now, for given AP,
First term, a = 31
And common difference, d = 28 – 31 = −3
As nth terms of an AP is given as,
            a_n = a + (n − 1)d
According to question,

Since, n comes out to be fraction but not a positive integer. Therefore, 0 cannot be a term of the given AP.
Question. 6 The taxi fare each km, when the fare is Rs. 15 for the first km and Rs. 8 for each additional km, does not form an AP as the total fare (in Rs.) After each km is 15, 8, 8, 8, …… . Is the statement true? Given Reasons.
Solution.
No, the statement is not ture as the total fare (in Rs.) after each km is given as:
            15, (15 + 8), (15 + 2 × 8), (15 + 3 × 8),… . = 15, 23, 31, 39,…
Let       t₁ = 15, t₂ = 23, t₃ = 31 and t₄ = 39
Now,   d₁ = t₂ – t₁ = 23 – 5 = 8
            d₂ = t₃ – t₂ = 33 – 23 = 8
And     d₃ ­= t₄ – t₃ = 39 – 31 = 8
As d₁ = d₂ = d₃, therefore given list of numbers formed by taking total fares after each Km forms an AP.
Question. 7 In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.
(i) The fee charged from a student every month by a school for the whole session, when the monthly is Rs. 400.
(ii) The fee charged every month by a school from classes I to XII, when the monthly fee for class I is Rs. 250 and it increase by Rs. 50 for the next higher class.
(iii) The amount of money in the account of Varun at the end of every year when Rs. 1000 is deposited at simple interest of 10% per annum.
(iv) The number of bacteria in certain food item after each second, when they double in every second.
Solution.
The fee charged from a student every month by a school for the whole session is
                                                400, 400, 400, 400,…
As the difference between successive terms of given list of numbers is same as 0, so it forms an AP, with common difference, d = 0.
(ii) The monthly fee from I to XII is given as:
            250, (250+50), (250 + 2 × 50), (250 + 3 × 50),….
i.e.,      250, 300, 350, 400,….
Here,   d₁ = 300 – 250 = 50
            d₂ = 350 – 300 = 50
            d₃ = 400 – 350 = 50
As, d₁ = d₂ = d₃, therefore given list of numbers formed by monthly fees from I to XII forms an AP.

So, the amount of money in the account of Varun at end of every year is given as:
            1000, (1000 + 100 × 1), (1000 + 100 × 2), (1000 + 100 × 3),…..
i.e.,      1000, 1100, 1200, 1300,….
Here,   d₁ = 1100 – 1000 = 100
            d₂ = 1200 – 1100 = 100
            d₃ = 1300 – 1200 = 100
As, d₁ = d₂ = d₃, therefore given list of numbers forms an AP.
(iv) Let the number of bacteria present in food initially = x
Since, they double in every second.
Therefore, number of bacteria after eavery second are given as:
            x, 2x, 2 (2x), 2 (2·2·x),…..
i.e.,      x, 2x, 4x, 8x,……
Here,   d₁ = 2x – x = x
            d₂ = 4x – 2x = 2x
            d₂ = 8x – 4x = 4x
As, d₁ = d₂ = d₃, therefore given list of numbers forms an AP.
Question. 8 Justify whether it is true to say that the following are the nth terms of an AP.
(i) 2n – 3
(ii) 3n² + 5
(iii) 1 + n + n²
Solution.
Concept used:To prove for the given term to be nth term of an AP, the series formed by that term must have same common difference.
(i) Yes.
Here, a_n = 2n – 3
⟹       a₁ = 2(1) – 3 = −1
            a₂ = 2(2) – 3 = 1
            a₃ = 2(3) – 3 = 3
            a₄ = 2(4) – 3 = 5
Now, d₁ = a₂ – a₁ = 1 – (–1) = 1 + 1 = 2
            d₂ = a₃ – a₂ = 3 – 1 = 2
            d₃ = a₄ – a₃ = 5 – 3 = 2
As, d₁ = d₂ = d₃, therefore list of numbers −1, 1, 3, 5,…. forms an AP.
Thus, 2n – 3 is the nth term of an AP.
(ii) No.
Here a_n = 3n² + 5
⟹       a₁ = 3(1)² + 5 = 8
⟹       a₂ = 3(2)² + 5 = 3(4) + 5 = 17
⟹       a₃ = 3(3)² + 5 = 3(9) + 5 = 27 + 5 = 32
⟹       a₄ = 3(4)² + 5 = 3(16) + 5 = 48 + 5 = 53
Now, d₁ = a₂ – a₁ = 17 – 8 = 9
d₂ = a₃ – a₂ = 32 – 17 = 15
d₃ = a₄ – a₃ = 53 – 32 = 21
As, d₁ ≠ d₂ ≠ d₃, therefore list of numbers 8, 17, 32, 53,……does not form an AP.
Thus, 3n² + 5 is not the nth term of an AP.
 (iii) No.
Here    a_n = 1 + n + n²
⟹       a₁ = 1 + 1 + (1)² = 3
⟹       a₂ = 1 + 2 + (2)² = 1 + 2 + 4 = 7
⟹       a₃ = 1 + 3 + (3)² = 1 + 3 + 9 = 13
⟹       a₄ = 1 + 3 + (4)² = 1 + 4 + 16 = 21
Now,   d₁ = a₂ – a₁ = 7 – 3 = 4
            d₂ = a₃ – a₂ = 13 – 7 = 6
            d₃ = a₄ – a₃ = 21 – 13 = 8                   
As, d₁ ≠ d₂ ≠ d₃, therefore list of numbers 3, 7, 13, 21,…. does not form an AP.
Thus, 1 + n + n² is not the nth term of an AP.
Exercise 5.3
Short Answer Type Questions
Question. 1 Match the AP’s given in column A with suitable common differences given in column B.

Solution.
(A₁) 2, -2, -6, -10,….
Here, common difference, d = – 2 – 2 = – 4
Hence, A₁ matches to B₄.
(A₂) a = −18, n = 10, a_n = 0
Using, a_n = a + (n – 1)d, we get:
            0 = −18 + (10 − 1)d
⟹       18 = 9d
⟹         d = 2
Hence, A₂ matches to B₅.
(A₃) a = 0, a₁₀ = 6

Hence, A₃ matches to B₁.
(A₄) a₂ = 13, a₄ = 3
Using, a_n = a + (n – 1)d, we get:
            a₂ = a + (2 − 1)d
⟹       a + d = 13                    …(i)
Again, a₄ = a + (4 − 1)d
⟹       a + 3d = 3                   ….(ii)
Subtracting equation (i) from equation (ii), we get:
            2d = −10
⟹          d = − 5
Hence, A₄ matches to B₂.
Question. 2 Verify that each of the following is an AP and Then Write its next three terms.

Solution.
A series of numbers will form an AP, if d₁ = d₂ = d₃…



(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), ….
Here, a₁ = a + b, a₂ = (a + 1) + b, a₃ = (a + 1) + (b + 1)
Now,   d₁ = a₂ – a₁ = [(a + 1) + b] – (a + b) = a + 1 + b – a – b = 1
          d₂ = a₃ – a₂ = [(a + 1) + (b + 1)] – [(a + 1) + b]
               = a + 1 + b + 1 – a – 1 – b = 1
As d₁ = d₂, therefore given list of numbers forms an AP.
Using, a_n = a + (n – 1)d, next three term are calculated as:
            a₄ = a₁ + 3d = a + b + 3(1) = (a+2) + (b+1)
            a₅ = a₁ + 4d = a + b + 4(1) = (a+2) + (b+2)
            a₆ = a₁ + 5d = a + b + 5(1) = (a+3) + (b+2)
(v)  a, 2a + 1, 3a + 2, 4a + 3, ….
Here, a₁ = a, a₂ = 2a + 1, a₃ = 3a + 2 and a₄ = 4a + 3
Now,   d₁ = a₂ – a₁ = 2a + 1 – a = a + 1
          d₂ = a₃ – a₂ = 3a + 2 – 2a – 1 = a + 1
          d₃ = a₄ – a₃ = 4a + 3 – 3a – 2 = a + 1
As d₁ = d₂ = d₃, therefore given list of numbers forms an AP.
Using, a_n = a + (n – 1)d, next three term are calculated as:
            a₅ = a₁ + 4d = a + 4 (a+1) = 5a + 4
            a₆ = a₁ + 5d = a + 5 (a+1) = 6a + 5
            a₇ = a₁ + 6d = a + 6 (a+1) = 7a + 6
Question. 3 Write the first three terms of the AP’s, When a and d are as given below

(ii) Given, first term, a = − 5
Common difference, d = − 3
Using, a_n = a + (n – 1)d, first three term of AP are calculated as:
Second term of AP, a₂ = a + d =  −5 – 3 =  −8
And third term of AP, a₃ = a + 2d = −5 + 2 (−3)
                                    = −5 – 6 = −11
Hence, required three terms are −5, −8, and −11.

Question. 4 Find a, b and c such that the following numbers are in AP:
                                     a, 7, b, 23 and c.
Solution.
Here,d₁ = 7 – a                     ….(i)
            d₂ = b – 7                    ….(ii)
            d₃ = 23 – b                  ….(iii)
            d₄ = c – 23                   ….(iv)
Since a, 7, b, 23 and c are in AP.
Therefore, d₁ = d₂ = d₃ = d₄
Taking (ii) and (iii), we get
            b − 7 = 23 – b
⟹       2b = 30
⟹       b = 15
Taking (i) and (ii), we get
            7 – a = b – 7
⟹       7 – a = 15 – 7             [Using b = 15]                        
⟹       7 – a = 8
⟹       a = − 1
Taking (iii) and (iv), we get
            23 – b = c – 23
⟹       23 – 15 = c – 23          [Using b = 15]                                    
⟹       8 = c – 23
⟹       8 + 23 = c
⟹       c = 31
Hence, a = − 1, b = 15, c = 31.
Question. 5 Determine the AP whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.
Solution.
Let the first term of AP be a and common difference d.
Using, a_n = a + (n – 1)d, we have:
            a₅ = a + (5 – 1) d = 19   [Given, a₅ = 19]
⟹       a + 4d = 19                  ….(i)
Also,    a₈ = a + (8 – 1) d = a + 7d
And,    a₁₃ = a + (13 – 1) d = a + 12d
According to question:
            a₁₃ − a₈ = 20
⟹       (a + 12d) – (a + 7d) = 20
⟹       5d = 20
⟹       d = 4
Putting d = 4 in equation (i), we get
            a + 4(4) = 19
⟹       a + 16 = 19
⟹       a = 19 – 16 = 3
So, required AP is given as: a, a + d, a + 2d, a + 3d, … = 3, 3 + 4, 3 + 2(4), 3 + 3(4), …= 3, 7, 11, 15, …..
Question. 6 The 26^th, 11^th and the last terms of an AP are, 0,3 and −1/5, respectively. Find the common difference and the number of terms.
Solution.
Let the first term, common difference and number of terms of AP be a, d and n, respectively.


Hence, the common difference and number of terms are – 1/ 5 and 27, respectively.
Question. 7 The sum of the 5th and the 7th terms of an AP is 52 and the 10^th term is 46. Find the AP.
Solution.
Let the first term and common difference of AP be a and d, respectively.
According to the question,
            a₅ + a₇ = 52 and a₁₀ = 46
⟹       (a + 4d) + (a + 6d) = 52   and   a + 9d = 46  [Using, a_n = a + (n – 1)d]
⟹       2a + 10d = 52 
Or        a + 5d = 26                  ….(i)
And     a + 9d = 46                  ….(ii)
Subtracting equation (i) from equation (ii), we get:
            4d = 20
⟹         d = 5
Putting d = 5, in equation (i) we get:
           a + 5 (5) = 26
⟹       a = 1
So, Required AP is given as: a, a + d, a + 2d, a + 3d, … = 1, 1 + 5, 1 + 2 (5), 1 + 3(5), …. = 1, 6, 11, 16, …..
Question. 8 Find the 20th term of the AP whose 7th term is 24 less than the 11th term, first term being 12.
Solution.
Let the first term, common difference and number of terms of an AP are a, d and n, respectively.
Given that, first term, a = 12.
According to question,
            a₇ = a₁₁ – 24    
⟹       a + 6d = (a + 10d) – 24           [Using, a_n = a + (n – 1)d]
⟹       4d = 24
⟹       d = 6
Hence, 20th term of the AP, a₂₀ = a + 19d = 12 + 19(6) = 126
Exercise 5.3
Short Answer Type Questions (Q. No. 20-29):
Question. 20 The first term of an AP is −5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Solution.
Let the first term, common difference and the number of terms of an AP are a, d and n respectively.
Given that, first term of the AP, a = −5
Last term of AP, l = 45
And, sum of all terms of AP, S_n = 120
We know that, if last term of an AP is known, then sum of n terms of an AP is,

Hence, number of terms and the common difference of an AP are 6 and 10 respectively.
Question. 21 Find the sum:

Solution.



Question. 22 Which term of the AP −2, −7, −12, … Will be −77? Find the sum of this AP upto the term −77.
Solution.
Let −77 be the nth term of AP −2, −7, −12,….
Here, first term, a = − 2
Common difference, d = −7 – (−2) = −7 + 2 = −5

Hence, the sum of the terms upto – 77 of given AP is − 632.
Question. 23 If a_n = 3 – 4n, then show that a₁, a₂, a₃, …. form an AP. Also, find S₂₀.
Solution.

Question. 24 In an AP, if S_n = n (4n + 1), then find the AP.
Solution.
Given, S_n = n (4n + 1)                         ….(i)
We know that, the nth term if an AP is given as:
            a_n = S_n − S_n−1
⟹       a_n = n (4n + 1) – (n – 1) [4(n – 1) + 1]            [Using (i)]
⟹       a_n = 4n² + n – (n – 1) (4n – 3)
⟹       a_n = 4n² + n – 4n² + 4n + 3n – 3
⟹       a_n = 8n − 3
Put n = 1, a₁ = 8(1) – 3 = 5
Put n = 2, a₂ = 8(2) – 3 = 16 – 3 = 13
Put n = 3, a₃ = 8(3) – 3 = 24 – 3 = 21
Hence, the required AP is 5, 13, 21, …..
Question. 25 In an AP, if S_n = 3n² + 5n and a_k = 164, then find the value of k.
Solution.
Given, S_n = 3n² + 5n               …(i)
And a_k = 164
We know that, the nth term if an AP is given as:
            a_n = S_n − S_n−1
            a_n = 3n² + 5n – 3(n −1)² – 5(n – 1)                  [Using (i)]
            a_n = 3n² + 5n – 3(n² + 1 – 2n) – 5n +5
            a_n = 3n² + 5n – 3n² – 3 +6n – 5n +5
            a_n =  6n + 2
Putting n = k in above equation and using equation (ii) , we get:
            a_k =  6k + 2 = 164      
⟹       6k = 162
⟹         k = 27
Question. 26 If S_n denotes the sum of first n terms of an AP, Then Prove that S₁₂ = 3 (S₈ – S₄).
Solution.

Question. 27 Find the sum of first 17 terms of an AP whose 4^th and 9^th terms are – 15 and – 30, respectively.
Solution.
Let the first term, common difference and the number of terms of AP be a, d and n, respectively.
Using formula for nth term of an AP, we have:
            a_n = a + (n − 1) d
⟹ 4^th term of AP, a₄ = a + 3d
⟹       a + 3d = −15               [Given]….(i)
Also, 9^th term of AP, a₉ = a+ 8d
⟹       a+ 8d = − 30 [Given]….(ii)    
Subtracting equation (ii) from equation (iii), We get

Hence, the required sum of first 17 terms of an AP is −510.
Question. 28 If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, then find the sum of first 10 terms.
Solution.
Let the first term and common difference of AP be a and d respectively.
Given that, sum of first 6 terms of AP, S₆ = 36
And, sum of first 16 terms of AP, S₁₆ = 256
As sum of first n terms of an AP is given as:

Question. 29 Find the sum of all the 11 terms of an AP whose middle most term is 30.
Solution.
Since the total number of terms, n = 11, which is odd.

Hence, the sum of all the 11 terms of AP is 330.
Exercise 5.4
Long Answer Type Questions (Q. No. 1-5):
Question. 1 The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the some of the first ten terms of this AP is 235, find the sum of its first twenty terms.
Solution.
Let the first term, common difference and the number of terms of given AP be a, d and n, respectively.



Question. 2 Find the
(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
(iii) Sum of those integers from 1 to 500 which are multiples of 2 or 5.
Solution.
(i) We first find the LCM of 2 and 5 which is 10.
Now all those integers which are multiples of 10 are also the multiples of 2 and 5.
Therefore, multiples of 2 as well as of 5 between 1 and 500 are 10, 20, 30, ……, 490 ….(i)
Series (i) forms an AP with first term, a = 10 and common difference, d = 20 – 10 = 10
Let total number of terms of this AP be n.
Therefore, nth term of AP, a_n = Last term, l = 490

(ii) All those integers which are multiples of 10 (LCM of 2 and 5) are also the multiples of 2 and 5.
Therefore multiples of 2 as well as of 5 from 1 to 500 are 10, 20, 30, ……., 500  ….(i)
Here, (i) forms an AP with,
First term, a = 10
Common difference, d = 10
Let total number of terms of this AP be n.
Therefore, nth term of AP, a_n = Last term, l = 490

(iii) Since, multiples of 2 or 5 from 1 to 500 = Multiple of 2 from 1 to 500 + Multiples of 5 from 1 to 500 – Multiple of 10 from 1 to 500
= (2, 4, 6, ….., 500) + (5, 10, 15, ….., 500) – (10, 20, ….., 500)          …..(i)
Let AP1 = 2, 4, 6, ….., 500
       AP₂ = 5, 10, 15, ….., 500
And AP₃ = 10, 20, ….., 500
Let number of terms in AP₁, AP₂ and AP₃ be n₁, n₂ and n₃ respectively.
Now formula for last term of an AP is given as:
            l = an + (n – 1)d                      ….(ii)              
 For AP₁,
            500 = 2 + (n₁ – 1)2      ….[Using (ii)]
⟹       250 = 1+ (n₁ – 1)
⟹         n₁ = 250
For AP₂,
            500 = 5 + (n₂ – 1)5      ….[Using (ii)]
⟹       100 = 1 + (n₂ – 1)
⟹          n₂ = 100
For AP₃,
            500 = 10 + (n₃ – 1)10  ….[Using (ii)]
⟹       250 = 1 + (n₃ − 1)
⟹       n₃ = 50
Thus, from equation (i),
Sum of Multiples of 2 or 5 from 1 to 500
            = Sum of (2, 4, 6, …… 500) + Sum of (5, 10, ….. 500) – Sum of (10, 20, ….. 500)

Question. 3 The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15^th term.
Solution.
Let the first term and common difference of AP be a and d, respectively.
According to the first condition, we have:


Question. 4 An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.
Solution.
Given, total number of terms, n = 37, which is odd.


Question. 5 Find the sum of the integers between 100 and 200 that are
(i) Divisible by 9.
(ii) Not divisible by 9.
Solution.
(i) The integers between 100 and 200 which is divisible by 9 are 108, 117, 126, … 198 which form an AP with first term, a = 108 and common difference,  d = 117 – 108 = 9
Let n be the number of terms between 100 and 200 which is divisible by 9.
Hence, required sum of the integers between 100 and 200 that are divisible by 9 is 1683.
(ii) The sum of the integers between 100 and 200 which is not divisible by 9 = (sum of total numbers between 100 and 200) – (Sum of total number between 100 and 200 which is divisible by 9).     ….(i)
For terms between 100 and 200,
First term, a = 101,  Common difference, d = 102 – 101 = 1  and a_n = l = 199

Using equation (i), Sum of the integers between 100 and 200 which
are not divisible by 9 = 14850 – 1683       [From solution of part (i)]
                               =13167
Exercise 5.4
Long Answer Type Questions (Q. No. 6-10):
Question. 6 The ratio of the 11^th term to the 18^th term of an AP is 2 : 3. Find the ratio of the 5^th term to the 21^st term and also the ratio of the sum of the first five terms to the sum of the first 21 terms.
Solution.
Let a and d be the first term and common difference of an AP.
Now nth term of an AP is given as:
            an = a + (n – 1)d
∴ 11^th term of AP, a₁₁ = a + 10d
And, 18^th term of AP, a₁₈ = a + 17d
According to question,


So, ratio of the sum of the first five terms of the AP to the sum of the first 21 terms is given as
            S₅ : S₂₁ = 30d  : 294d = 5 : 49
Question. 7 Show that the sum of an AP whose first term is a, the second term b and the last


Hence proved.
Question. 8 Solve the equation – 4 + (–1) + 2 + … + x = 437.
Solution.
Given equation is, – 4 – 1 + 2 + …. + x = 437                                    …. (i)
LHS of equation (i) forms an AP with
First term, a = – 4,
Common difference, d = 3,
And last term, l = a_n  = x


Here, x cannot be negative as for x = −53, n will be negative which is not possible
Hence, the required value of x is 50.
Question. 9 Jaspal singh repays his total loan of Rs. 118000 by paying every month starting with the first installment of Rs. 1000. If he increases the installment by Rs. 100 every month, what amount will be paid by him in the 30^th installment? What amount of loan does he still have to pay after the 30^th installment?
Solution.
Jaspal singh takes total loan = Rs. 118000
Monthly instalments paid by Jaspal are Rs. 1000, Rs. 1100, Rs. 1200, …. Which form an AP with:
First term, a = 1000
Common difference, d = 100
30^th term of this AP, a₃₀ = 1000 + (30 − 1)100           [Using, a_n = a + (n−1)d]
                                    = 1000 + 29 × 100
                                    = 1000 + 2900 = 3900
So, Rs. 3900 will be paid by Jaspal in the 30^th installment.
Now, Total amount paid in 30 installments is = 1000 + 1100 + 1200 + ….. + 3900
Here, First term, a = 1000,
Last term, l = 3900
And, number of terms, n = 30

Total amount that Jaspal still have to pay after the 30^th installment
            = (Amount of loan) – (Sum of 30 installments)
            = 118000 – 73500 = Rs. 44500
Question. 10 The students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags.
Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance she did cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?
Solution.

Thus Ruchi have to place 13 flags on the left of 14th flag and 13 flags on its right.
Similarly, for placing third flag and return his initial position, distance travelled = 4+ 4= 8m
For placing fourth flag and return his initial position, distance travelled = 6+ 6 = 12 m.
For placing fourteenth flag and return his initial position, distance travelled
                                                = 26 + 26 = 52m
Proceed same manner into her right position from middle most flag i.e., 14th flag.
Distance between two adjacent flags = 2m
Therefore distance travelled to place first flag = 4m
Distance travelled to place second flag = 8m
Distance travelled to place 13^th flag = 52 m
Also, when Ruchi placed the last flag she returns his middle most position and collects her books. This distance also included in placed the last flag.
So, these distances form a series.
            4 + 8+ 12 + 16 + …. + 52        [For left side]
And     4 + 8 + 12 + 16 + ….. + 52      [For right side]
Here, first term, a = 4
Common difference, d = 8 – 4 = 4
And, number of terms, n = 13

Now, the maximum distance she travelled carrying a flag = Distance travelled by Ruchi during placing the 14th flag in her left position or 27th flag in her right position
            = (2 + 2 + 2 + … + 13 times)
            = 2 × 13 = 26 m
Hence, the maximum distance she travelled carrying a flag is 26 m.