5

Q1. The sum of the first n even numbers is 

• 1) n^₂ + n
• 2) n^₂ - 1
• 3) n^₂
• 4) n^₂ + 2

Solution

Q2. 228 logs are to be stacked in a store in the following manner : 30 logs in the bottom, 28 in the next row, then 26 and so on. In how many rows can these 228 logs be stacked? How many logs are there in the last row?

Solution

Logs are stacked in the following manner, 30, 28, 26 .... a = 30,d = 28 - 30 = -2 Sn = [2a + (n - 1) d] 228 = [60 + (n - 1) (-2)] 456 = n [60 - 2n + 2] 456 = n [62 - 2n] 456 = -2n² + 62n 2n² - 62n + 456 = 0 n² - 31n + 228 = 0 n² - 19n - 12n + 228 = 0 n (n - 19) - 12 (n - 19) = 0 (n - 12) (n - 19) = 0 n = 12, 19 If the last row be 19^th number of logs in the last row will be= a + (n - 1)d = 30 + 18 (-2) = 30 - 36 = -6 Since the number of logs cannot be negative, the number of rows should be 12. Last row will have = 30+(12-1)x(-2)=30 + 11 (-2) = 8 logs

Q3. Find the sum of the first 10 natural numbers. 

• 1) 56
• 2) 57
• 3) 55
• 4) 54

Solution

The series is 1,2,3,… 10.  

Q4. Check whether    forms an Arithmetic  progression or not .If it is an AP then write the common difference and the next two  terms.

Solution

 

Q5. A class X student made plans for his study schedule for the month of February 2008.He decided to increase the number  study hours everyday by quarter of an hour.If he started with one hour on the first of February how many hours of studies canhe put in by the end of February 2008?

Solution

Here we essentially need to find the sum of an A.P.  with first term a=1, the common difference d=0.25 hours (quarter of an hour) And n=29 (as the year 2008 will have 29 days in the month of February) We know that for an A.P. with first term as ‘a’ and common diference as ‘d’ the sum of first n terms is given by   So the student can put in 130.5 hours of studies by the end of February 2008

Q6. Find the sum of the following arithmetic series: −2, −4, −6,…, −100 

• 1) −1254
• 2) 1254
• 3) −2550
• 4) 2550

Solution

Here, a = −2 and d = −2 and a_n = −100 ⇒ a + (n − 1)d = −100 ⇒ (−2) + (n − 1) × (−2) = −100 ⇒ −2− 2n +2 = −100 ⇒ −2n = −100 ⇒ n = 50 Therefore, the required sum of the series = [2 × −2 + (50− 1) × −2] = 25[−4 −98] = 25 × −102 = −2550

Q7. The sum of five consecutive numbers is 100. Find the first number.

• 1) 16
• 2) 17
• 3) 18
• 4) 19

Solution

5 consecutive numbers form an arithmetic progression with difference 1.n = 5,S₅ = 100,d = 1Let the first number be a. It is unknown.

Q8. Which term of the AP 3, 8, 13, 18,… will be 55 more than its 20^th term? 

• 1) 31
• 2) 30
• 3) 32
• 4) 29

Solution

3, 8, 13, 18,…  a = 3, d = 8 - 3 = 5 According to the question,T_n = 55 + T₂₀ ∴T_n= 55 + a + 19d ∴T_n = 55 + 3 + 19 × 5 ∴T_n = 153 ∴a + (n - 1)d = 153 ∴3 + (n - 1) × 5 = 153 ∴5(n - 1) = 150 ∴n - 1 = 30 ∴n = 31

Q9. There is  a race , in which, every participant starts from a bucket, picks up the nearest ball and runs back to drop it in the bucket. Then again, he or she runs back to nearest ball and runs back to drop it in the bucket. Each participant continues like this in the same way until all the balls are in the bucket. The balls are kept at a distance of 4 m from each other and there are 8 balls in all.  How much distance does each participant needs to travel in order to finish the race?

Solution

Distance travelled by the participant in order to pick up the first ball and drop it in the bucket= 4 + 4 = 8 m Distance travelled by the participant to pick up the second ball and drop it in basket = 4 + 4 + 4 + 4 = 16 m Similarly, distance travelled by competitor to pick up third, fourth, fifth, sixth, seventh, and eighth balls are 24 m, 32 m, 40 m, 48 m, 56 m, 64 m     a = 8 m common difference  = d = 8m       So, each participant has to cover a distance of 288 m in order to complete the race

Q10. A manufacturer of radio sets produced 600 units in the 3^rd year and 700 units in the 7^th year. Assuming that the production uniformly increases by fixed number every year. Find (a) the production in the 1^st year (b) the total production in 7 years (c) the production in the 10^th year.

Solution

Given that the production uniformly increases by fixed number every year. So, the production in each year forms an AP. Let the AP be a, a + d, a + 2d, …… Production in the third year = a₃ = a + 2d = 600… (1) Production in the seventh year = a₇ = a + 6d = 700 … (2) Subtracting (1) from (2), we get: 4d = 100 d = 25 _{From (1), we get:} a = 600 - 50 = 550 Thus, production in the first year is 550 units. Total production in 7 years = S₇ = = =  = 4375 Thus, the total production in 7 years is 4375 units. Production in the 10^th year = a₁₀ = a + 9d = 550 + 9 25 = 775 units

Q11. Find the sum of the following arithmetic series: 2 + 4 + 6+ … to 10 terms 

• 1) 130
• 2) 110
• 3) 120
• 4) 100

Solution

The sum of the first n terms of the AP whose first term = a and common difference = d is given by

Q12. The next term of the A.P. ... is:

• 1)
• 2)
• 3)
• 4)

Solution

Common difference =

Q13. Find the sum of first twelve multiples of 7.

Solution

Q14. Sum of the first n terms of an A.P. is 5n² - 3n. Find the A.P. and also find its 16^th term.

Solution

Sum of the n terms of an A.P. is 5n² - 3n. Find the terms of the A.P. and also find the 16^th term. S_n = 5n² - 3n a_n = S_n - S_{n - 1} a_n = 5n² - 3n - [5(n - 1)² - 3 (n - 1)] a_n = 5n² - 3n -[5(n² - 2n + 1) - 3n + 3] a_n = 5n² - 3n - 5n² + 10n - 5 + 3n - 3 a_n = 10n - 8 a₁ = 2 a₂ = 20 - 8 = 12 a₃ = 30 - 8 = 22 a₁₆ = 160-8= 152

Q15. What is the common difference of the A.P. in which a₁₈­ - a₁₄ = 32?

• 1) 4
• 2) 8
• 3) - 4
• 4) - 8

Solution

We know a_n = a + (n - 1)d, where a is the first term and d is the common difference. a₁₈ - a₁₄ = 32 a + (18 - 1)d - [a + (14 - 1)d] = 32 17d - 13d = 32 d =   d = 8 Common difference is 8.

Q16. Find the sum of all 15 terms of an AP whose middle-most term is 56.

• 1) 610
• 2) 840
• 3) 480
• 4) 410

Solution

LetA = first term and d = common difference of the given AP. In an AP consisting of 15 terms, i.e. 8^th term is the middle term. Given: middle term = 56 ∴ a + 7d = 56 … (i) We know that  

Q17. If the first term of an AP is 252 and the common difference is 3, then the last term is 999 and the number of terms is 250, then what is the sum of the AP? 

• 1) 156375
• 2) 146375
• 3) 166375
• 4) 186375

Solution

Q18. Find the sum of 1+ 4 + 7 + 10 + 13 + …40. 

• 1) 262
• 2) 166
• 3) 160
• 4) 287

Solution

Let there be n terms in the givenarithmetic series, then  a_n= 40 ⇒ a + (n − 1)d = 40 ⇒ 1 + (n − 1) × 3 = 40 ⇒ 1 + 3n − 3 = 40 ⇒ 3n − 2 = 40 ⇒ 3n = 42 ⇒ n = 14 Therefore, the required sum of the series = [2 × 1 + (14− 1) × 3] = 7 [2 + 13 × 3] = 7 [2 + 39] = 7 × 41 = 287

Q19. 15^th term of the A.P. x - 7, x - 2, x + 3 ... is:

• 1) x + 83
• 2) x + 53
• 3) x + 73
• 4) x + 63

Solution

a₁ = x - 7 a₂ = x - 2 common difference = d = x - 2 - (x - 7) = 5 15^th term: a₁₅ = a₁ + (n - 1)d a₁₅ = x - 7 + (15 - 1) x 5     = x + 63

Q20. Find the sum of the AP −26, −24, −22,…to the 36th term. 

• 1) 123
• 2) 144
• 3) 322
• 4) 324

Solution

Here, a = −26 and d = −24 − (−26) = −24 + 26 = 2

Q21. Find the common difference of an A.P. whose first term is  and the 8^th term is . Also, write its 4^th term.

Solution

t₈ = a + 7d    

Q22. Find three numbers in A.P. whose sum is 15 and the product is 80.

Solution

25 - d² = 16 d² = 9 d = 3 _{When d = 3, the numbers are 2, 5, 8 When d = -3, the numbers are 8, 5, 2}

Q23. Which of the following is the formula for the sum of the first n natural numbers? 

• 1)
• 2)
• 3)
• 4)

Solution

Sum of the first n natural numbers = 

Q24. Write the next two terms of the following AP:  3,  1,  -1,  -3,  ……..?

Solution

A. P is 3, 1,  -1,  - 3……           Here           Common difference          

Q25. The common difference of an A.P. in which a₂₅ - a₁₂ = - 52 is:

• 1) - 4
• 2) 3
• 3) - 3
• 4) 4

Solution

  We know a_n = a + (n - 1)d, where a is the first term and d is the common difference. a₂₅ - a₁₂ = - 52 a + (25 - 1)d - [a + (12 - 1)d] = - 52 24d - 11d = - 52 d =

Q26. Which term of the A.P. 1, 4, 7 ... is 88?

• 1) 35
• 2) 27
• 3) 30
• 4) 26

Solution

The given sequence is 1, 4, 7, … This is an AP with first term (a) = 1 and common difference (d) = 3 Let a_n = 88 a + (n - 1)d = 88 1 + (n - 1)(3) = 88 3(n - 1) = 87 n = 29 + 1 = 30 Thus, 88 is the 30^th term of the given sequence.

Q27. The sum of the third and seventh term of an A.P. is 6 and their product is 8. Find the sum of the first sixteen terms of the A.P.

Solution

Let the A.P. be a, a + d, a + 2d, ...... t₃ + t₇ = 6 2a + 8d = 6 a + 4d = 3 a = 3 - 4d… (1)   (a + 2d) (a + 6d) = 8 Using (1), we get, _{(3 - 2d) (3 + 2d) = 8 9 - 4d² = 8} If If

Q28. A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than it preceding prize, find the value of each of the prizes.

Solution

Let the value of the prices be x, x - 20, x - 40, ..... This is an arithmetic sequence with first term (a) = x and common difference (d) = -20. Given,  S₇ = 700 _{Sn =  [2a + (n - 1)d]} Thus, the values of the prizes are Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs, 60, and Rs. 40.

Q29. In an AP, if the 5^th and 12^th terms are 30 and 65, respectively, what is the sum of the first 20 terms? 

• 1) 1150
• 2) 2348
• 3) 4556
• 4) 2345

Solution

Here, a₅ = a + 4d = 30 and a₁₂ = a + 11d = 65 a + 4d = 30 … (i) a + 11d = 65 … (ii) Solving both equations, we get a = 10 and d = 5

Q30. Find the 10^th term from the end of the A.P.series 4,9,14, ..... 254.

Solution

AP in reverse order : 254, ..... , 14,9,4 a = 254 d = 4 - 9 = -5 a₁₀ = a + (10 - 1)d = 254 + 9 x (-5) = 209 10^th term from the end of given AP = 209

Q31. Find the sum of all multiples of 9 lying between 300 and 700.

Solution

The multiples of 9 lying between 300 and 700 are 306, 315, ..... 693 This is an AP with first term, a = 306 and common difference, d = 9

Q32. Find the sum of the following arithmetic series: 7, 10, 13,… to 14 terms 

• 1) 371
• 2) 120
• 3) 230
• 4) 234

Solution

Q33. An organization has new officers for the year and is ordering new stationery. In January, a mailing is sent to the 136 current members. If the membership increases each month by 6 members, how many envelopes will be needed for one year’s monthly mailings?

Solution

   In January, number of envelopes sent = 136  In February, number of envelopes needed = 142  In March, number of envelopes needed = 148  In April, number of envelopes needed = 154  In May, number of envelopes needed = 160 and so on…………………..   The number of envelopes required form an arithmetic progression with the first term (a) = 136 and common difference = 142 – 136 = 6 We will have to find the sum of the first 12 terms of our arithmetic progression in this case to find the total number of envelopes needed for one year’s monthly mailings.                                         Hence, total required envelopes in a year = 2028  

Q34. A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months, his savings increase by Rs. 40 more than the savings of the immediately previous month. His total savings from the start of service will be Rs. 11040 after how many months? 

• 1) 28
• 2) 21
• 3) 32
• 4) 30

Solution

According to the question, the series is 200, 200, 200, 240, 280, 320,…  Sum of the series = 11040 200 + 200 + 200 + 240 + 280 + 320,… = 11040 200 + 240 + 280 + 320,…. = 11040 - 400 200 + 240 + 280 + 320,…. = 10640 a = 200, d = 40 ∴  ∴n(400 + 40n - 40) = 21280 ∴n(40n + 360) = 21280 ∴40n(n + 9) = 21280 ∴n(n + 9) = 532 ∴n(n + 9) = 28 × 19 ∴n = 19 Hence, his total savings from the start of service will be Rs. 11040 after 19+2 = 21 months.

Q35. If mtimesthe m^thterm of an AP is equal to n times the n^thterm and m ≠ n, then (m + n)^th term is

• 1) None of these
• 2) m
• 3) n
• 4) m + n

Solution

According to the question, T_m = n a + (n - 1)d = n … (i)T_n = m a + (m - 1)d = m … (ii)mT_m= n T_n ∴ m[a + (m - 1)d] = n [a + (n - 1)d] ∴ ma - na + m (m - 1)d - n (n - 1)d = 0 ∴ a (m - n) + m² - md - n² + nd = 0 ∴ a (m - n) + (m - n)(m + n) - (m - n)d = 0 ∴ (m - n)[a + (m + n - 1)d] = 0 ∴ a + (m + n - 1)d = 0 ∴ T_{m + n} = 0

Q36. Find the sum of the first 20 terms of an AP in which the first term is −1 and the third term is7. 

• 1) 563
• 2) 654
• 3) 345
• 4) 740

Solution

Here, a = −1 and a₃ = 7 ∴a + 2d = 7 ∴−1 + 2d = 7 ∴ d = 4

Q37. Find the sum of the first 30 positive integers divisible by 3 and 2 both.

Solution

The numbers which are divisible by by 3 and 2 both are the numbers that are divisible by 6. So first 30 such numbers would be forming an A.P.with 6 as the fist term and 6 as the common difference So we should find the sum 6+12+18+… to 30 terms   We know that for an A.P. with first term as ‘a’ and common diference as ‘d’ the sum of first n terms is given by = So in this case we get, =     =2790

Q38. Which term of the A.P. 92, 88, 84, 80 ... is 0?

• 1) 24
• 2) 32
• 3) 23
• 4) 22

Solution

  Let the n^th of the given AP be 0. We know that, the n^th term of an AP is given by a + (n - 1)d, where a is its first term and d is its common difference. Here, a = 92 and d = 88 - 92 = - 4 0 = 92 + (n - 1) (- 4)    0 = 92 - 4n + 4  4n = 96    n = 24 Thus, the 24^th term of the given AP is 0.

Q39. The first term of an A.P. is 5 and its 100^th term is –292.  Find the 50^th term of this A.P.

Solution

Let the A.P. be             Here,                         a₁ = 5                                                           We have                                                                                                                                                 d = –3             Also                                                                                                                                                        50^th term is –142.  

Q40. Which term of the A.P. 21, 42, 63, 84, …… is 420 ?

Solution

Given A.P. is 21, 42, 63, 84, ….. 420                                                      

Q41. Find the  terms of the sequence defined by          

Solution

For n = 5 which is an odd natural number we have                     For n = 8 which is an even natural number            a_n = n(n+1) = 8(8+1) = 8 x 9 = 72

Q42. If p - 1, p + 3, 3p - 1 are in A.P., then p is equal to:

• 1) - 4
• 2) 4
• 3) 2
• 4) - 2

Solution

p - 1, p + 3, 3p - 1 are in A.P. p + 3 - p + 1 = 3p - 1 - p - 3 4 = 2p - 4 2p = 8 p = 4

Q43. Find the sum of the given AP: 3, 9, 15,…99, where the number of terms is 17. 

• 1) 7336
• 2) 4326
• 3) 5049
• 4) 6216

Solution

Q44. Write the next two terms of the following sequence          

Solution

          First term                    

Q45. Find the sum of 30 terms of the AP 1, 4, 7, 10,… 

• 1) 1335
• 2) 1219
• 3) 1191
• 4) 1129

Solution

Q46. Which of the following is the formula for the sum of the n terms of an AP? 

• 1)
• 2)
• 3)
• 4)

Solution

Q47. In November 2009, the number of visitors to a zoo increased daily by 20. If a total of 12300 people visited the zoo in that month, find the number of visitors on 1^st Nov. 2009.

Solution

Let number of visitors in zoo on 1^st November be x Then the daily visitors in November in the zoo are: x, x + 20 ... This will form an AP with first term x and common difference 20. Total no. of visitors in Nov. = 12300 (Given) S₃₀ = 12300 12300 = 15 (2x + 29 20) = 30 x + 8700 x = 120 Number of visitors on 1^st Nov. 2009 = 120.

Q48. 8^th term of an A.P. is 37 and its 12^th term is 57. Find the A.P.

Solution

_{Let a be the first term and d be the common difference of the AP. We know that the nth term of an AP is given by:} Thus, the A.P. is 2, 7, 12, ...

Q49. How many terms of the A.P. 78, 71, 64, ..... are needed to give the sum 465? Also find the last term of this A.P.

Solution

a = 78, d = 71 - 78 = -7 Let n be the required no. of terms S_n = 465 [2a + (n - 1)d] = 465 n[156 + (n - 1) (-7)] = 930 7n ² - 163n + 930 = 0 (7n - 93) (n - 10) = 0 n = or n = 10 Neglect n = as n cannot be a fraction No. of terms (n) = 10

Q50. The 4^th term of an A.P. is equal to 3 times the first term and the 7^th term exceeds twice the 3^rd term by 1. Find the A.P.

Solution

Let a and d respectively be the first term and the common difference of the AP. n^th term = a_n = a + (n - 1)d Given, a₄ = 3a _{From (1), 2a = 6  a = 3} A.P. is 3, 5, 7, ...

Q51. A sum of Rs. 700 is to be used for giving 7 cash prizes to students of a school for their academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.

Solution

The value of prizes will form an AP with common difference (d) = -20. We know that the sum of n terms of an AP is given by: S_n = It is given that S₇ = 700 700 = 200 = 2a - 120 a = 160 Thus, the cash prizes are 160, 140, 120, 100, 80, 60, 40

Q52. In an A.P., prove that a_{m + n} + a_{m - n} = 2 a_m, where a_n denotes nth term of the A.P.

Solution

We know that n^th term of an AP is given by a_n = a + (n - 1)d. a_{m + n} = a + (m + n - 1) d… (i) a_{m - n} = a + (m - n - 1) d… (ii) a_m = a + (m - 1) d… (iii) Adding (i) and (ii), = 2a + 2d (m - 1) = 2 [a + (m - 1)d] = 2a_m

Q53. Vikram invested a certain amount in shares. The stock market, in that year, reached new heights throughout. The stock market grew at a rate such that his investment grew by a fixed amount every month. As a result, after four months, his investment became Rs. 10,000 and after nine months, it became Rs. 17,500. What amount will Vikram have after the twelfth month?

Solution

Let the initial amount invested by Vikram be Rs.a. Let his investment increase by Rs.d each month. Initial investment = a Thus, investment after the 1^st month = a + d Investment after the 2^nd month = a + 2d. Investment after the n^th month = a + nd   Investment after the 4^th month = a + 4d = 10,000  … (1) Investment after the 9^th month = a + 9d = 17,500  … (2)   Subtracting (1) from (2): 9d – 4d = 17,500 – 10,000 5d = 7,500 d = 1500   Substituting the value of d in (1): a + 4(1500) = 10,000 a = 10,000 – 6000 a = 4000   Thus, he invested Rs.4,000 in the stock market and his investment grew by Rs.1500 each month. Thus, after the 12^th month, his investment became a + 12d = Rs. 4,000 + 12 x (1500) = Rs. 4,000 + 18,000 = Rs. 22,000

Q54.

Solution

Q55. The 4^th term of an AP is equal to 3 times the first term and the 7^th term exceeds twice the 3^rd term by 1. Find the first term and the common difference.

Solution

Q56. In a flower bed, there are 31 rose plants in the first row, 28 in the second, 25 in the third, and so on. There are 7 rose plants in the last row. How many rows are there in the flower bed?

Solution

The number of rose plants in the 1st, 2nd, 3rd … rows are 31, 28, 25 … 7                           These numbers are in an AP. Let the number of rows in the flower bed be n. Then a = 31, d = 28 – 31 = – 3, a_n = 7                                                 a_n = a + (n – 1) d                                                                       Thus, 7 = 31 + (n – 1)(– 3) i.e., – 24 = (n – 1)(– 3) i.e.  8 = n - 1 i.e., n = 9 So, there are 9 rows in the flower bed. 

Q57.

Solution

Q58. A spiral is made up of successive semicircles, with centres alternately at A and B, starting from A, of radius 0.5 cm, 1 cm, ..... What is the total length of such a spiral made up of thirteen consecutive semicircles?

Solution

Circumference of a semi-circle = = r Total length = r₁ + r₂ + r₃ + ... = (0.5 + 1 + 1.5 + ..... upto 13 terms) =

Q59. Kartik repays his total loan of Rs. 1,18,000 by paying every month starting with the first instalment of 1000, if he increases the instalment by Rs. 100 every month. What amount will be paid by him in the 30^th instalment? What amount of loan does he still have to pay after the 30^th instalment?

Solution

Total amount of loan = Rs 1,18,000 Since the amount of each installment increases by Rs 100 every month, therefore, the installments paid are in A.P. Amount of first installment, a = Rs 1000 Increase in amount of each installment, d = Rs 100 Amount to be paid in 30^th instalment = a₃₀ a₃₀ = a + (n - 1)d = 1000 + 29 100 = 1000 + 2900 = Rs. 3900 Amount of loan to be still paid after paying the 30^th installment = Total amount of loan - Total amount paid by him in 30 installments Amount paid in 30 instalments = S₃₀ S₃₀ = [2a + (n - 1)d] = 15[2000 + 2900] = 15 4900 = 73,500 Amount of loan to be still paid after paying the 30^th installment = Rs 1,18,000 - Rs 73,500 = Rs 44,500

Q60. Find the sum of the first 50 odd natural numbers.

Solution

50 odd natural numbers are 1,3,5,...a_n Sn = n/2[2a+(n-1)d]= n/2(2x1 + 49x2) = 50/2 (100) = 2500

Q61. How many terms of the A.P. 9,17,25, ..., must be taken to get a sum of 450?

Solution

a = 9, d = 8, S_n = 450 S_n = [2a + (n - 1) d] 450 = [18 + (n - 1) (8)] 450 = 4n² + 5n 4n² + 5n - 450 = 0 4n² - 40n + 45n - 450 = 0 4n (n - 10) + 45 (n - 10) = 0 (n - 10) (4n + 45) = 0 n = or n = 10 Rejecting n =  as number of terms can not be negative. n = 10 Thus, ten terms of the given A.P. will make sum as 450.

Q62. Find the sum of all three digit numbers which leave the same remainder 2 when divided by 5.

Solution

The three digit numbers which leave the same remainder 2 when divided by 5 are 102, 107, ..... 997 This is an AP. First term, a = 102 Common difference, d = 5 Last term, a_n = 997 We know that the nth term of an AP is given by: a_n = a + (n - 1)d S_n = [a + l] = [102 + 997] = 90 1099 = 98910 Hence, the required sum of all three digit numbers which leave the same remainder 2 when divided by 5 is 98910.

Q63. Which term of the A.P. 3, 15, 27, 39, ... will be 132 more than its 60^th term?

Solution

First term = a = 3 Common difference = d = 15 - 3 = 12 We know that the n^th term of an AP is given by: a_n = a + (n - 1)d a₆₀­ = 3 + (60 - 1) 12 = 3 + 708 = 711 _{Now, we need to find which term is 132 more than 60^th term. 132 + 711 = 843}

Q64. Find the sum of all two-digit odd positive numbers.

Solution

Two digit odd positive numbers are 11, 13, 15, ...., 99. This is an A.P. with, First term, a = 11, Common difference, d = 13 - 11 = 2, and last term, a_n = 99 Now, a_n = a + (n - 1)d 99 = 11 + (n - 1)2 99 = 11 + 2n - 2 99 = 9 + 2n 2n = 90 n = 45 Thus, the sum of all two-digit odd positive numbers is 2475.

Q65. Find the number of terms of the series: -5 + (-8) + (-11) + ...... + (-230)

Solution

-5 + (-8) + (-11) + ...... + (-230) The series -5, -8, -11, … -230 is an arithmetic progression with First term, a = -5 Common difference, d = -3 Let a_n = -230 -230 = a + (n - 1)d -230 = -5 + (n - 1)(-3) -230 = -5 - 3n + 3 -228 = -3n n = 76 Thus, there are 76 terms in the given sequence.

Q66. Which term of the Arithmetic Progression 3, 10, 17,…. Will be 84 more than its 13^th term?

Solution

Q67. Find the sum of the first 31 terms of an A.P. whose n^th term is given by 3 +  n.

Solution

= _{Thus, the sum of the first 31 terms of the A.P. is  .}

Q68. The sum of the first three terms of an A.P. is 33. If the product of the first and the third term exceeds the second term by 29, find the A.P.

Solution

Let the terms in AP be (a - d), a, (a + d). Sum of the three terms = a - d + a + a + d = 3a 3a = 33 a = 11 Also, from the given information, we have: (a - d)(a + d) = a + 29 a² - d² = 11 + 29 = 40 121 - d² = 40 d² = 81 d = 9 Thus, the AP is 2, 11, 20, … or 20, 11, 2, …

Q69. Calculate how many multiples of 7 are there between 100 and 300.

Solution

Multiples of 7 between 100 and 300 are 105, 112, ... 294 First term = a = 105 Common difference = d = 7 Let 294 be the n^th term of the AP. Then, 294 = a + (n - 1)d 294 = 105 + (n - 1)7 7(n - 1) = 189 n = 28

Q70. A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follow: Rs.200 for 1^st day, Rs.250 for second day, Rs.300 for third day, and so on. If the contractor pays Rs.27750 as penalty, find the number of days for which the construction work was delayed.

Solution

Penalty for 1^st day = Rs.200 Penalty for 2^nd day = Rs.250 Penalty for 3^rd day = Rs.300; and so on. Total penalty = Rs.27750 This penalty is an A.P. with a = 200,  d (common difference) = 50            Let the work be completed after n days                                        S_n = 27750                            Since the number of days cannot be negative, n = 30 days