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Algebra · Textbook solutions

Arithmetic Progression

Every solved example, exercise, and miscellaneous question, in the order the textbook teaches them. Plus every board question asked from this chapter. · 92 questions

Arithmetic Progression

8 q

Solved Examples

Worked · 8
  1. Which of the following sequences are A.P ? If it is an A.P, find next two terms.
    Arithmetic Progression SolvedEx.1 (i)
    5, 12, 19, 26, . . .
  2. Arithmetic Progression SolvedEx.1 (ii)
    2, -2, -6, -10, . . .
  3. Arithmetic Progression SolvedEx.1 (iii)
    1, 1, 2, 2, 3, 3, . . .
  4. Arithmetic Progression SolvedEx.1 (iv)
    32\dfrac{3}{2}, 12\dfrac{1}{2}, −12-\dfrac{1}{2}, . . .
  5. The first term aa and common difference dd are given. Find first four terms of A.P.
    Arithmetic Progression SolvedEx.2 (i)
    a=−3a = -3, d=4d = 4
  6. Arithmetic Progression SolvedEx.2 (ii)
    a=200a = 200, d=7d = 7
  7. Arithmetic Progression SolvedEx.2 (iii)
    a=−1a = -1, d=−12d = -\dfrac{1}{2}
  8. Arithmetic Progression SolvedEx.2 (iv)
    a=8a = 8, d=−5d = -5

Practice set 3.1

18 q
  1. Which of the following sequences are A.P. ? If they are A.P. find the common difference .
    Ex 3.1 Q.1(1)
    2, 4, 6, 8, . . .
  2. Ex 3.1 Q.1(2)
    2, 52\dfrac{5}{2}, 3, 73\dfrac{7}{3}, . . .
  3. Ex 3.1 Q.1(3)
    -10, -6, -2, 2, . . .
  4. Ex 3.1 Q.1(4)
    0.3, 0.33, .0333, . . .
  5. Ex 3.1 Q.1(5)
    0, -4, -8, -12, . . .
  6. Ex 3.1 Q.1(6)
    −15-\dfrac{1}{5}, −15-\dfrac{1}{5}, −15-\dfrac{1}{5}, . . .
  7. Ex 3.1 Q.1(7)
    3, 3+23 + \sqrt{2}, 3+223 + 2\sqrt{2}, 3+323 + 3\sqrt{2}, . . .
  8. Ex 3.1 Q.1(8)
    127, 132, 137, . . .
  9. Write an A.P. whose first term is aa and common difference is dd in each of the following.
    Ex 3.1 Q.2(1)
    a=10a = 10, d=5d = 5
  10. Ex 3.1 Q.2(2)
    a=−3a = -3, d=0d = 0
  11. Ex 3.1 Q.2(3)
    a=−7a = -7, d=12d = \dfrac{1}{2}
  12. Ex 3.1 Q.2(4)
    a=−1.25a = -1.25, d=3d = 3
  13. Ex 3.1 Q.2(5)
    a=6a = 6, d=−3d = -3
  14. Ex 3.1 Q.2(6)
    a=−19a = -19, d=−4d = -4
  15. Find the first term and common difference for each of the A.P.
    Ex 3.1 Q.3(1)
    5, 1, -3, -7, . . .
  16. Ex 3.1 Q.3(2)
    0.6, 0.9, 1.2, 1.5, . . .
  17. Ex 3.1 Q.3(3)
    127, 135, 143, 151, . . .
  18. Ex 3.1 Q.3(4)
    14\dfrac{1}{4}, 34\dfrac{3}{4}, 54\dfrac{5}{4}, 74\dfrac{7}{4}, . . .

nth term of an A. P.

5 q

Solved Examples

Worked · 5
  1. nth term of an A.P. SolvedEx.1
    Find tnt_n for following A.P. and then find 30th30^{\text{th}} term of A.P. 3, 8, 13, 18, . . .
  2. nth term of an A.P. SolvedEx.2
    Which term of the following A.P. is 560 ? 2, 11, 20, 29, . . .
  3. nth term of an A.P. SolvedEx.3
    Check whether 301 is in the sequence 5, 11, 17, 23, . . . ?
  4. nth term of an A.P. SolvedEx.4
    How many two digit numbers are divisible by 4 ?
  5. nth term of an A.P. SolvedEx.5
    The 10th10^{\text{th}} term and the 18th18^{\text{th}} term of an A.P. are 25 and 41 respectively then find 38th38^{\text{th}} term of that A.P., similarly if nthn^{\text{th}} term is 99. Find the value of nn.

Practice set 3.2

13 q
  1. Write the correct number in the given boxes from the following A. P.
    Ex 3.2 Q.1(i)
    1, 8, 15, 22, . . . Here a=□a = \square, t1=□t_1 = \square, t2=□t_2 = \square, t3=□t_3 = \square, t2−t1=□−□=□t_2 - t_1 = \square - \square = \square t3−t2=□−□=□t_3 - t_2 = \square - \square = \square ∴\therefore d=□d = \square
  2. Ex 3.2 Q.1(ii)
    3, 6, 9, 12, . . . Here t1=□t_1 = \square, t2=□t_2 = \square, t3=□t_3 = \square, t4=□t_4 = \square, t2−t1=□t_2 - t_1 = \square, t3−t2=□t_3 - t_2 = \square ∴\therefore d=□d = \square
  3. Ex 3.2 Q.1(iii)
    -3, -8, -13, -18, . . . Here t3=□t_3 = \square, t2=□t_2 = \square, t4=□t_4 = \square, t1=□t_1 = \square, t2−t1=□t_2 - t_1 = \square, t3−t2=□t_3 - t_2 = \square ∴\therefore a=□a = \square, d=□d = \square
  4. Ex 3.2 Q.1(iv)
    70, 60, 50, 40, . . . Here t1=□t_1 = \square, t2=□t_2 = \square, t3=□t_3 = \square, . . . ∴\therefore a=□a = \square, d=□d = \square
  5. Ex 3.2 Q.2
    Decide whether following sequence is an A.P., if so find the 20th20^{\text{th}} term of the progression. -12, -5, 2, 9, 16, 23, 30, . . .
  6. Ex 3.2 Q.3
    Given Arithmetic Progression 12, 16, 20, 24, . . . Find the 24th24^{\text{th}} term of this progression.
  7. Ex 3.2 Q.4
    Find the 19th19^{\text{th}} term of the following A.P. 7, 13, 19, 25, . . .
  8. Ex 3.2 Q.5
    Find the 27th27^{\text{th}} term of the following A.P. 9, 4, -1, -6, -11, . . .
  9. Ex 3.2 Q.6
    Find how many three digit natural numbers are divisible by 5.
  10. Ex 3.2 Q.7
    The 11th11^{\text{th}} term and the 21st21^{\text{st}} term of an A.P. are 16 and 29 respectively, then find the 41th41^{\text{th}} term of that A.P.
  11. Ex 3.2 Q.8
    11, 8, 5, 2, . . . In this A.P. which term is number -151 ?
  12. Ex 3.2 Q.9
    In the natural numbers from 10 to 250, how many are divisible by 4 ?
  13. Ex 3.2 Q.10
    In an A.P. 17th17^{\text{th}} term is 7 more than its 10th10^{\text{th}} term. Find the common difference.

Sum of first n terms of an A. P.

4 q

Solved Examples

Worked · 4
  1. Sum of first n terms of an A.P. SolvedEx.1
    Find the sum of first nn natural numbers.
  2. Sum of first n terms of an A.P. SolvedEx.2
    Find the sum of first nn even natural numbers.
  3. Sum of first n terms of an A.P. SolvedEx.3
    Find the sum of first nn odd natural numbers.
  4. Sum of first n terms of an A.P. SolvedEx.4
    Find the sum of all odd numbers from 1 to 150.

Practice set 3.3

9 q
  1. Ex 3.3 Q.1
    First term and common difference of an A.P. are 6 and 3 respectively ; find S27S_{27}. a=6a = 6, d=3d = 3, S27=?S_{27} = ? Sn=n2[□+(n−1)d]S_n = \dfrac{n}{2}[\square + (n - 1)d] S27=272[12+(27−1)□]S_{27} = \dfrac{27}{2}[12 + (27 - 1)\square] =272×□= \dfrac{27}{2} \times \square =27×45=□= 27 \times 45 = \square
  2. Ex 3.3 Q.2
    Find the sum of first 123 even natural numbers.
  3. Ex 3.3 Q.3
    Find the sum of all even numbers between 1 and 350.
  4. Ex 3.3 Q.4
    In an A.P. 19th19^{\text{th}} term is 52 and 38th38^{\text{th}} term is 128, find sum of first 56 terms.
  5. Ex 3.3 Q.5
    Complete the following activity to find the sum of natural numbers between 1 and 140 which are divisible by 4. Between 1 and 140, natural numbers divisible by 4 4, 8, . . . . . . . . , 136 How many numbers ? ∴\therefore n=□n = \square n=□n = \square, a=□a = \square, d=□d = \square tn=a+(n−1)dt_n = a + (n - 1)d 136=□+(n−1)×□136 = \square + (n - 1) \times \square n=□n = \square ⟶\longrightarrow Sn=n2[2a+(n−1)d]S_n = \dfrac{n}{2}[2a + (n - 1)d] S□=□2[    ]=□S_{\square} = \dfrac{\square}{2}[\ \ \ \ ] = \square Sum of numbers from 1 to 140, which are divisible by 4 =□= \square
  6. Ex 3.3 Q.6
    Sum of first 55 terms in an A.P. is 3300, find its 28th28^{\text{th}} term.
  7. Ex 3.3 Q.7
    In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms. (Assume that three consecutive terms in A.P. are a−da - d, aa, a+da + d.)
  8. Ex 3.3 Q.8
    Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd3^{\text{rd}} and 4th4^{\text{th}} term is 14. (Assume the four consecutive terms in A.P. are a−da - d, aa, a+da + d, a+2da + 2d.)
  9. Ex 3.3 Q.9
    If the 9th9^{\text{th}} term of an A.P. is zero then show that the 29th29^{\text{th}} term is twice the 19th19^{\text{th}} term.

Application of A.P.

6 q

Solved Examples

Worked · 6
  1. Application of A.P. SolvedEx.1
    A mixer manufacturing company manufactured 600 mixers in 3rd3^{\text{rd}} year and in 7th7^{\text{th}} year they manufactured 700 mixers. If every year there is same growth in the production of mixers then find (i) Production in the first year (ii) Production in 10th10^{\text{th}} year (iii) Total production in first seven years.
  2. Application of A.P. SolvedEx.2
    Ajay sharma repays the borrowed amount of Rs. 3,25,000\text{Rs. }3{,}25{,}000 by paying Rs. 30500\text{Rs. }30500 in the first month and then decreases the payment by Rs. 1500\text{Rs. }1500 every month. How long will it take to clear his amount ?
  3. Application of A.P. SolvedEx.3
    Anvar saves some amount every month. In first three months he saves Rs. 200\text{Rs. }200, Rs. 250\text{Rs. }250 and Rs. 300\text{Rs. }300 respectively. In which month will he save Rs. 1000\text{Rs. }1000 ?
  4. Application of A.P. SolvedEx.4
    As shown in the figure, take point A on the line and draw a half circle P1P_1 of radius 0.5 with A as centre. It intersects given line in point B. Now taking B as centre draw a half circle P2P_2 of radius 1 cm which is on the other side of the line. Now again taking A as centre draw a half circle P3P_3 of radius 1.5 cm. If we draw half circles like this having radius 0.5 cm, 1 cm, 1.5 cm, 2 cm, we get a figure of spiral shape. Find the length of such spiral shaped figure formed by 13 such half circles. (π=227)\left(\pi = \dfrac{22}{7}\right)
  5. Application of A.P. SolvedEx.5
    In the year 2010 in the village there were 4000 people who were literate. Every year the number of literate people increases by 400. How many people will be literate in the year 2020 ?
  6. Application of A.P. SolvedEx.6
    In year 2015, Mrs. Shaikh got a job with salary Rs. 1,80,000\text{Rs. }1{,}80{,}000 per year. Her employer agreed to give Rs. 10,000\text{Rs. }10{,}000 per year as increment. Then in how many years will her annual salary be Rs. 2,50,000\text{Rs. }2{,}50{,}000 ?

Practice set 3.4

6 q
  1. Ex 3.4 Q.1
    On 1st1^{\text{st}} Jan 2016, Sanika decides to save Rs. 10\text{Rs. }10, Rs. 11\text{Rs. }11 on second day, Rs. 12\text{Rs. }12 on third day. If she decides to save like this, then on 31st31^{\text{st}} Dec 2016 what would be her total saving ?
  2. Ex 3.4 Q.2
    A man borrows Rs. 8000\text{Rs. }8000 and agrees to repay with a total interest of Rs. 1360\text{Rs. }1360 in 12 monthly instalments. Each instalment being less than the preceding one by Rs. 40\text{Rs. }40. Find the amount of the first and last instalment.
  3. Ex 3.4 Q.3
    Sachin invested in a national saving certificate scheme. In the first year he invested Rs. 5000\text{Rs. }5000, in the second year Rs. 7000\text{Rs. }7000, in the third year Rs. 9000\text{Rs. }9000 and so on. Find the total amount that he invested in 12 years.
  4. Ex 3.4 Q.4
    There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th15^{\text{th}} row and also find how many total seats are there in the auditorium ?
  5. Ex 3.4 Q.5
    Kargil's temperature was recorded in a week from Monday to Saturday. All readings were in A.P.The sum of temperatures of Monday and Saturday was 5∘5^\circ C more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was −30∘-30^\circ celsius then find the temperature on the other five days.
  6. Ex 3.4 Q.6
    On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.

Problem set 3

23 q
  1. Choose the correct alternative answer for each of the following sub questions.
    PS3 Q.1(1)
    The sequence -10, -6, -2, 2, . . .

    Tap an option to check your answer.

  2. PS3 Q.1(2)
    First four terms of an A.P. are ....., whose first term is -2 and common difference is -2.

    Tap an option to check your answer.

  3. PS3 Q.1(3)
    What is the sum of the first 30 natural numbers ?

    Tap an option to check your answer.

  4. PS3 Q.1(4)
    For an given A.P. t7=4t_7 = 4, d=−4d = -4 then a=…a = \ldots

    Tap an option to check your answer.

  5. PS3 Q.1(5)
    For an given A.P. a=3.5a = 3.5, d=0d = 0, n=101n = 101, then tn=…t_n = \ldots

    Tap an option to check your answer.

  6. PS3 Q.1(6)
    In an A.P. first two terms are -3, 4 then 21st21^{\text{st}} term is . . .

    Tap an option to check your answer.

  7. PS3 Q.1(7)
    If for any A.P. d=5d = 5 then t18−t13=…t_{18} - t_{13} = \ldots

    Tap an option to check your answer.

  8. PS3 Q.1(8)
    Sum of first five multuiples of 3 is. . .

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  9. PS3 Q.1(9)
    15, 10, 5, . . . In this A.P. sum of first 10 terms is . . .

    Tap an option to check your answer.

  10. PS3 Q.1(10)
    In an A.P. 1st1^{\text{st}} term is 1 and the last term is 20. The sum of all terms is = 399 then n=…n = \ldots

    Tap an option to check your answer.

  11. PS3 Q.2
    Find the fourth term from the end in an A.P. -11, -8, -5, . . . , 49.
  12. PS3 Q.3
    In an A.P. the 10th10^{\text{th}} term is 46, sum of the 5th5^{\text{th}} and 7th7^{\text{th}} term is 52. Find the A.P.
  13. PS3 Q.4
    The A.P. in which 4th4^{\text{th}} term is -15 and 9th9^{\text{th}} term is -30. Find the sum of the first 10 numbers.
  14. PS3 Q.5
    Two A.P.'s are given 9, 7, 5, . . . and 24, 21, 18, . . . . If nthn^{\text{th}} term of both the progressions are equal then find the value of nn and nthn^{\text{th}} term.
  15. PS3 Q.6
    If sum of 3rd3^{\text{rd}} and 8th8^{\text{th}} terms of an A.P. is 7 and sum of 7th7^{\text{th}} and 14th14^{\text{th}} terms is -3 then find the 10th10^{\text{th}} term.
  16. PS3 Q.7
    In an A.P. the first term is -5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there ? What is the common difference ?
  17. PS3 Q.8
    Sum of 1 to nn natural numbers is 36, then find the value of nn.
  18. PS3 Q.9
    Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
  19. PS3 Q.10
    There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.
  20. PS3 Q.11
    If first term of an A.P. is aa, second term is bb and last term is cc, then show that sum of all terms is (a+c)(b+c−2a)2(b−a)\dfrac{(a + c)(b + c - 2a)}{2(b - a)}.
  21. PS3 Q.12
    If the sum of first pp terms of an A.P. is equal to the sum of first qq terms then show that the sum of its first (p+q)(p + q) terms is zero. (p≠q)(p \neq q)
  22. PS3 Q.13
    If mm times the mthm^{\text{th}} term of an A.P. is eqaul to nn times nthn^{\text{th}} term then show that the (m+n)th(m + n)^{\text{th}} term of the A.P. is zero.
  23. PS3 Q.14
    Rs. 1000\text{Rs. }1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity. Simple interest =P×R×N100= \dfrac{P \times R \times N}{100} Simple interest after 1 year =1000×10×1100=□= \dfrac{1000 \times 10 \times 1}{100} = \square Simple interest after 2 year =1000×10×2100=□= \dfrac{1000 \times 10 \times 2}{100} = \square Simple interest after 3 year =□×□×□100=300= \dfrac{\square \times \square \times \square}{100} = 300 According to this the simple interest for 4, 5, 6 years will be 400, □\square, □\square respectively. From this d=□d = \square, and a=□a = \square Amount of simple interest after 20 years tn=a+(n−1)dt_n = a + (n - 1)d t20=□+(20−1)□t_{20} = \square + (20 - 1)\square t20=□t_{20} = \square Amount of simple interest after 20 years is =□= \square