CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 24

Exercise 1: Subjective Questions

Question 1. Is 51 a term of the A.P. 5, 8, 11, 14, ...?
Answer:
a = 5, d = 3. \( 51 = 5 + (n - 1)3 \Rightarrow n = \dfrac{49}{3} \), not a whole number. No.
In simple words: Solve a_n = 51.

Teacher's Note:
1. n must be a natural number.

Question 2. Which term of the A.P. 8, 14, 20, 26, ... will be 72 more than its 41st term?
Answer:
\( a_n = a_{41} + 72 \Rightarrow (n - 41)d = 72 \Rightarrow (n - 41) \times 6 = 72 \Rightarrow n = \) 53. It is the 53rd term.
In simple words: The extra 72 is a multiple of d.

Teacher's Note:
1. d = 6.

Question 3. Write the next term of \( \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \)
Answer:
The terms are \( 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2} \). The next term is \( 5\sqrt{2} = \) \( \sqrt{50} \).
In simple words: Simplify the surds.

Teacher's Note:
1. d = √2.

Question 4. If \( \dfrac{4}{5}, a, 2 \) are three consecutive terms of an A.P., find the value of a.
Answer:
\( 2a = \dfrac{4}{5} + 2 = \dfrac{14}{5} \Rightarrow a = \) \( \dfrac{7}{5} \).
In simple words: The middle term is the average.

Teacher's Note:
1. The source prints no answer; derived here.

Question 5. Find the 10th term from the end of the A.P. 4, 9, 14, ..., 254.
Answer:
\( 254 = 4 + (n - 1)5 \Rightarrow n = 51 \). The 10th term from the end is the \( (51 - 10 + 1) = 42 \)nd term: \( 4 + 41 \times 5 = \) 209.
In simple words: Count from the end: m-th from end is (n - m + 1)th.

Teacher's Note:
1. Or reverse the A.P.: 254, 249, ... with d = -5.

Question 6. The fourth term of an A.P. is equal to three times the first term and the seventh term exceeds twice the third term by one. Find the first term and the common difference.
Answer:
\( a + 3d = 3a \Rightarrow 3d = 2a \). \( a + 6d = 2(a + 2d) + 1 \Rightarrow 2d = a + 1 \). Then \( \dfrac{4a}{3} = a + 1 \Rightarrow a = 3 \), \( d = 2 \). First term 3, common difference 2.
In simple words: Make two equations.

Teacher's Note:
1. Check: a4 = 9 = 3 x 3.

Question 7. Which term of the sequence \( 17, 16\tfrac{1}{5}, 15\tfrac{2}{5}, 14\tfrac{3}{5}, \dots \) is the first negative term?
Answer:
\( d = -\dfrac{4}{5} \). \( 17 - \dfrac{4}{5}(n - 1) < 0 \Rightarrow n - 1 > 21.25 \Rightarrow n = 23 \). The 23rd term is the first negative term.
In simple words: Solve a_n < 0.

Teacher's Note:
1. a_22 = 0.2 and a_23 = -0.6.

Question 8. Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21st term?
Answer:
\( (n - 21)d = 120 \Rightarrow (n - 21) \times 12 = 120 \Rightarrow n = \) 31.
In simple words: Divide 120 by d.

Teacher's Note:
1. d = 12.

Question 9. Find the common difference of an A.P. whose first term is 100 and the sum of whose first six terms is five times the sum of the next six terms.
Answer:
\( S_6 = 5(S_{12} - S_6) \Rightarrow 6S_6 = 5S_{12} \). \( 6(600 + 15d) = 5(1200 + 66d) \Rightarrow 3600 + 90d = 6000 + 330d \Rightarrow \) d = -10.
In simple words: The next six terms sum to S_12 - S_6.

Teacher's Note:
1. a = 100.

Question 10. Find three numbers in A.P. whose sum is 21 and their product is 336.
Answer:
Let the numbers be a - d, a, a + d: \( 3a = 21 \Rightarrow a = 7 \). \( 7(49 - d^2) = 336 \Rightarrow d^2 = 1 \). The numbers are 6, 7, 8 (or 8, 7, 6).
In simple words: Use the a - d, a, a + d form.

Teacher's Note:
1. d = ±1 gives the same numbers.

Question 11. A student purchased a pen for Rs. 100. At the end of 8 years, it was valued at Rs. 20. Assuming that the yearly depreciation is a constant amount, find the annual depreciation.
Answer:
The values form an A.P.: 100 at the start and 20 after 8 years: \( 20 = 100 + 8d \Rightarrow d = -10 \). The annual depreciation is Rs. 10.
In simple words: The value falls by the same amount each year.

Teacher's Note:
1. 8 steps from 100 to 20.

Question 12. Find the sum of all the three digit numbers which leave remainder 2 when divided by 5.
Answer:
The numbers are 102, 107, ..., 997. \( 997 = 102 + (n - 1)5 \Rightarrow n = 180 \). \( S = 90(102 + 997) = \) 98910.
In simple words: They form an A.P. with d = 5.

Teacher's Note:
1. The first is 100 + 2 = 102.

Question 13. Find the sum of all two-digit odd positive numbers.
Answer:
11, 13, ..., 99: \( 99 = 11 + (n - 1)2 \Rightarrow n = 45 \). \( S = \dfrac{45}{2}(11 + 99) = \) 2475.
In simple words: Odd numbers form an A.P. with d = 2.

Teacher's Note:
1. Two-digit odd numbers start at 11.

Question 14. The sum of the first n terms of an A.P. is given by \( S_n = 3n^2 - 4n \). Determine the A.P. and its 12th term.
Answer:
\( a_1 = S_1 = -1 \), \( a_2 = S_2 - S_1 = 4 + 1 = 5 \). d = 6. The A.P. is -1, 5, 11, ... and \( a_{12} = -1 + 66 = \) 65.
In simple words: a_n = S_n - S_(n-1).

Teacher's Note:
1. General: a_n = 6n - 7.

Question 15. Find the sum of the first 25 terms of an A.P. whose nth term is given by \( t_n = 2 - 3n \).
Answer:
\( t_1 = -1 \), \( t_{25} = -73 \). \( S_{25} = \dfrac{25}{2}(-1 - 73) = \) -925.
In simple words: Use S = n(first + last)/2.

Teacher's Note:
1. d = -3.

Question 16. In an A.P., the sum of the first n terms is \( \dfrac{3n^2}{2} + \dfrac{5n}{2} \). Find its 25th term.
Answer:
\( S_{25} = \dfrac{3 \times 625 + 125}{2} = 1000 \) and \( S_{24} = \dfrac{3 \times 576 + 120}{2} = 924 \). \( a_{25} = 1000 - 924 = \) 76.
In simple words: a_25 = S_25 - S_24.

Teacher's Note:
1. General: a_n = 3n + 1.

Question 17. In an A.P., if \( S_n = n^2 p \) and \( S_m = m^2 p \) \( (m \ne n) \), prove that \( S_p = p^3 \).
Answer:
From \( S_n = pn^2 \) the nth term is \( a_n = pn^2 - p(n - 1)^2 = p(2n - 1) \) (a first term p and common difference 2p). Hence \( S_p = p \cdot p^2 = p^3 \) (put n = p in \( S_n = pn^2 \)).
In simple words: The formula S_n = pn² holds for every n.

Teacher's Note:
1. Substituting n = p gives the result.

Question 18. Find the number of terms of the A.P. 54, 51, 48, ... so that their sum is 513.
Answer:
\( \dfrac{n}{2}[108 - 3(n - 1)] = 513 \Rightarrow n^2 - 37n + 342 = 0 \Rightarrow (n - 18)(n - 19) = 0 \). n = 18 or 19. (The 19th term is 0, so both give the same sum.)
In simple words: Solve the quadratic in n.

Teacher's Note:
1. a_19 = 54 - 54 = 0.

Question 19. The first term, common difference and last term of an A.P. are 12, 6 and 252 respectively. Find the sum of all its terms.
Answer:
\( 252 = 12 + (n - 1)6 \Rightarrow n = 41 \). \( S = \dfrac{41}{2}(12 + 252) = \) 5412.
In simple words: Find n first.

Teacher's Note:
1. 264/2 = 132, and 41 x 132 = 5412.

Question 20. The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
Answer:
\( 2a + 10d = 24 \) and \( 2a + 14d = 44 \Rightarrow 4d = 20 \Rightarrow d = 5 \), \( a = -13 \). The first three terms are -13, -8, -3.
In simple words: Subtract the two equations.

Teacher's Note:
1. Check: a4 = 2 and a8 = 22.

Question 21. The sum of the first six terms of an A.P. is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and the thirteenth term of the A.P.
Answer:
\( 3(a + 9d) = a + 29d \Rightarrow a = d \). \( S_6 = 3(2a + 5d) = 21d = 42 \Rightarrow d = 2, a = 2 \). \( a_{13} = 2 + 24 = \) 26. First term 2, thirteenth term 26.
In simple words: The ratio gives a = d.

Teacher's Note:
1. Check: a10 = 20 and a30 = 60.

Question 22. If the sum of the first q terms of an A.P. is \( 2q + 3q^2 \), what is its common difference?
Answer:
\( S_q = 3q^2 + 2q \), so \( a_q = S_q - S_{q-1} = 6q - 1 \). The common difference is 6.
In simple words: a_q - a_(q-1) is constant.

Teacher's Note:
1. The coefficient of q² is d/2.

Question 23. Find the common difference of an A.P. whose first term is 4, the last term is 49 and the sum of all its terms is 265.
Answer:
\( 265 = \dfrac{n}{2}(4 + 49) \Rightarrow n = 10 \). \( 49 = 4 + 9d \Rightarrow d = \) 5.
In simple words: Find n from the sum formula.

Teacher's Note:
1. Check: 4, 9, ..., 49.

Question 24. In an A.P., the sum of its first ten terms is -80 and the sum of its next ten terms is -280. Find the A.P.
Answer:
\( S_{10} = 10a + 45d = -80 \) and \( S_{20} = -360 = 20a + 190d \). Subtracting twice the first equation: \( 100d = -200 \Rightarrow d = -2 \), \( a = 1 \). The A.P. is 1, -1, -3, -5, ...
In simple words: S_20 = -80 + (-280).

Teacher's Note:
1. Check S_10 = 5(2 - 18) = -80.

Word Problems

Question 25. 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?
Answer:
\( \dfrac{n}{2}[40 - (n - 1)] = 200 \Rightarrow n^2 - 41n + 400 = 0 \Rightarrow (n - 16)(n - 25) = 0 \). n = 25 would need a negative number of logs in the top row (20 - 24 = -4), so n = 16 rows and the top row has \( 20 - 15 = \) 5 logs.
In simple words: Reject the root that gives a negative last term.

Teacher's Note:
1. a_25 would be -4, impossible.

Question 26. Jaspal Singh repays his loan of Rs. 118000 by paying every month starting with the first instalment of Rs. 1000. If he increases the instalment by Rs. 100 every month, what amount will be paid by him in the 30th instalment? What loan does he still have to pay after the 30th instalment?
Answer:
a = 1000, d = 100. \( a_{30} = 1000 + 29 \times 100 = \) Rs. 3900. \( S_{30} = 15(1000 + 3900) = 73500 \). Loan left \( = 118000 - 73500 = \) Rs. 44500.
In simple words: The instalments form an A.P.

Teacher's Note:
1. Subtract the sum paid from the loan.

Question 27. 150 workers were engaged to finish a work in a certain number of days. Four workers were dropped on the second day, four more workers were dropped on the third day and so on. It takes 8 more days to finish the work now. Find the total number of days to finish the work.
Answer:
Let the work be planned for n days: total work \( = 150n \) worker-days. Now it takes \( n + 8 \) days with workers 150, 146, 142, ... (an A.P. with d = -4). \( \dfrac{n + 8}{2}[300 - 4(n + 7)] = 150n \Rightarrow (n + 8)(136 - 2n) = 150n \Rightarrow n^2 + 15n - 544 = 0 \Rightarrow (n - 17)(n + 32) = 0 \). n = 17, so the work now takes \( 17 + 8 = \) 25 days.
In simple words: Total work stays the same.

Teacher's Note:
1. Check: sum of 150, 146, ..., 54 = 25 x 102 = 2550 = 150 x 17.

Question 28. Two cars start simultaneously from the same place and in the same direction. The first car moves with a uniform speed of 10 km/h. The second car moves at a speed of 8 km/h in the first hour and then increases its speed by 0.5 km/h in each succeeding hour. If both the cars move non-stop, how long will it take for the second car to overtake the first car?
Answer:
After n hours the second car has gone \( \dfrac{n}{2}[16 + 0.5(n - 1)] \) km and the first car 10n km. Setting them equal: \( 16 + 0.5(n - 1) = 20 \Rightarrow n = 9 \). The second car overtakes the first after 9 hours.
In simple words: Equal distances at the overtaking time.

Teacher's Note:
1. Divide both sides by n/2.