CBSE Class 10 Mathematics chapter 5 - ARITHMETIC PROGRESSIONS - Exercise 5.2

 CBSE Class 10 Mathematics chapter 5 - ARITHMETIC PROGRESSIONS -  Exercise 5.2

Q1.	Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:
	a d n an (i) 7 3 8 … (ii) —18 … 10 0 (iii) … —3 18 —5 (iv) —18.9 2.5 … 3.6 (v) 3.5 0 105 …
Solution.	(i) Given = a = 7, d = 3, n = 8 an = a + (n—1)d Putting value an = 7 + (8—1)×3 an = 7 + 7 × 3 an = 7 + 21 an = 28 (ii) Given = a = —18, n = 10, an = 0 an = a + (n—1)d d = an—a/n—1 d = 0—(—18)/10—1 d = 18/9 = 2 (iii) Given = d = —3, n = 18, an = —5 an = a + (n—1)d a = an—(n—1)d a = (—5)—(18—1)(—3) a = (—5)—(17)(—3) a = —5 + 51 a = 46 (iv) Given = a = —18.9, d = 2.5, an = 3.6 an = a + (n—1)d n = 1 + (an—a)/d n = 1+(3.6—(—18.9))/2.5 n = 1+(3.6+18.9)/2.5 n = 1+22.5/2.5 n = 1 + 9 = 10 (v) Given = a = 3.5, d = 0, n = 105 an = a + (n—1)d Putting value an = 3.5 + (105—1)×0 an = 3.5 + 104 × 0 an = 3.5 + 0 an = 3.5

Q2.	Choose the correct choice in the following and justify :
(i)	30th term of the AP: 10, 7, 4, . . . , is (A) 97                    (B) 77                  (C) –77                 (D) – 87
Solution.	Option (C) is correct. Explanation: Given AP: 10, 7, 4, … We have to fine 30th term, so n = 30 a = 10 d = 7—10 = —3 We know that an = a + (n—1)d Putting value an = 10 + (30—1)×(—3) an = 10 + 29×(—3) an = 10—87 an = —77
(ii)	11th term of the AP: – 3, –½ , 2, . . ., is (A) 28                    (B) 22                  (C) –38                 (D) – 48½
Solution.	Option (B) is correct. Explanation: Given AP: – 3, –½ , 2, . . . We have to fine 11th term, so n = 11 a = —3 d = (—½)—(—3) = —½ + 3 = —2.5 We know that an = a + (n—1)d Putting value an = (—3) + (11—1)×(—2.5) an = (—3) + 10×2.5 an = —3+25 an = 22

Q3.	In the following APs, find the missing terms:
(i)	2, … , 26
Solution.	Let missing term = x Then AP become = 2, x, 26 We know that – d = a2 – a1 = a3 – a2 Common difference = second term — First term = Third term — second term Putting value x — 2 = 26 — x x + x = 26 + 2 2x = 28 x = 14 So, Missing term is 14.

(ii)

…, 13,  …,  3

Solution.

Here given term is a2 = 13 and a4 = 3 We know that – an = a + (n—1)d if a2 is 13, then 13 = a + (2—1)d 13 = a + 1×d 13 = a + d a = 13 — d …………. (i) if a4 is 3, then 3 = a + (4—1)d 3 = a + 3×d 3 = a + 3d a = 3 —3d …………… (ii) by equation (i) and (ii) 13—d = 3—3d 13—3 = d—3d 10 = —2d d = —5 putting value in equation (i) a = 13—(—5) a = 18 For find a3 – an = a + (n—1)d a3 = 18 + (3—1)(—5) a3 = 18 + (2)(—5) a3 = 18 — 10 a3 = 8 So, missing term is 18 and 8.

(iii)

5, …,  …,  9½

Solution.

1st Method : Here First term (a) = 5 and last term(l) = 9½ For finding common difference: d = (Last term —First term) / (Number of term —1) Putting value – d = (9½ — 5)/(4—1) d = 4½/3 d = 1½ We know that an = a + (n—1)d If n = 2 a2 = 5 + (2—1)×1½ a2 = 5 + 1×1½ a2 = 5 + 1½ a2 = 6½ If n = 3 a3 = a2 + d a3 = 6½ + 1½ a3 = 8 So, missing term is 6½ and 8. Second Method : Given a = 5 a4 = 9½ = 9.5 We know that d = an—a/n—1 d = 9.5—5/4—1 d = 4.5/3 d = 1.5 an = a + d a2 = 5 + 1.5 = 6.5 a3 = 6.5 + 1.5 = 8

(iv)

—4, …, …, …, …, 6

Solution.

1st Method : Here First term (a) = —4 and last term(l) = 6 For finding common difference: d = (Last term —First term) / (Number of term —1) Putting value – d = (6 — (—4)/(6—1) d = (6 +4)/(6—1) d = 10/5 d = 2 We know that an = a + (n—1)d If n = 2 a2 = (—4) + (2—1)×2 a2 = —4 + 1×2 a2 = —4 + 2 a2 = —2 If n = 3 a3 = a2 + d a3 = —2 + 2 a3 = 0 If n = 4 a4 = a3 + d a4 = 0 + 2 a4 = 2 If n = 5 a5 = a4 + d a5 = 2 + 2 a5 = 4 So, missing term is —2, 0, 2 and 4. Second Method : Given a = —4 a6 = 6 We know that d = an—a/n—1 d = 6—(—4)/6—1 d = 10/5 d = 2 an = a, a + d, a + 2d, a + 3d, a + 4d, … a2 = (—4) + 2 = —2 a3 = (—4) + 2(2) = 0 a4 = (—4) + 3(2) = 2 a5 = (—4) + 4(2) = 4 So, missing term is —2, 0, 2 and 4.

(v)

…, 38, …, …, …, —22

Solution.

1st Method : Here second term (a2) = 38,  last term(l) = —22 and n = 6 We know that an = a + (n—1)d a = an — (n—1)d if an = a2 a = 38—(2—1)d a = 38—1×d a = 38—d ……………… (i) if an = a6 a = (—22)—(6—1)d a = —22 — 5 × d a = —22—5d ……………… (ii) from equation (i) and (ii) 38—d = —22—5d 38+22 = d—5d 60 = —4d d = —15 put value in equation (i) a = 38—(—15) a = 38+15 a = 53 For find a3,a4 and a5 Now we have a = 53 and d  = —15 We know that an = a, a + d, a + 2d, a + 3d, a + 4d, … a3 = 53 + 2(—15) = 23 a4 = 53 + 3(—15) = 8 a5 = 53 + 4(—15) = —7 So, missing term is 53, 23, 8 and —7.

Q4.	Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
Solution.	Given – a = 3 d = 8—3 = 5 an = 78 We know that – an = a + (n—1)d Putting value 78 = 3 + (n—1) 5 78—3 = (n—1) 5 75 = (n—1) 5 75/5 = n—1 15 = n—1 15+1 = n n = 16 So, 78 is 16th term of given AP.

Q5.	Find the number of terms in each of the following APs :
(i)	7, 13, 19, . . . , 205
Solution.	Given – a = 7 d = 13—7 = 6 an = 205 We know that – an = a + (n—1)d Putting value 205 = 7 + (n—1) 6 205—7 = (n—1) 6 198 = (n—1) 6 198/6 = n—1 33 = n—1 33+1 = n n = 34 So, there are 34th terms in given AP.

(ii)	18, 15½ , 13, . . . , – 47
Solution.	Given – a = 18 d = 15.5—18 = 2.5         (because we can write 15½ = 15.5) an = —47 We know that – an = a + (n—1)d Putting value —47 = 18 + (n—1) 2.5 —47—18 = (n—1) 2.5 65 = (n—1) 2.5 65/2.5 = n—1 26 = n—1 26+1 = n n = 27 So, there are 27th terms in given AP.

Q6.	Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .
Solution.	Given – a = 11 d = 8—11 = —3 an = —150 (Assume) We know that – an = a + (n—1)d Putting value —150 = 11 + (n—1) (—3) —150—11 = (n—1) (—3) 161 = (n—1) (—3) 161/3 = n—1 53.66 = n—1 53.66 +1 = n n = 54.66 So, n present as decimal form, which is wrong. Hence, —150 is not a term of given AP.

Q7.

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution.

Given – a11 = 38 a16 = 73 a31 = ? We know that – an = a + (n—1)d For a11 term - Putting value 38 = a + (11—1) d 38 = a + 10d ……………….. (i) For a16 term - Putting value 73 = a + (16—1) d 73 = a + 15d ……………….. (ii) From equation (i) and (ii) (a + 15d) — (a + 10d) = 73 — 38 a + 15d — a — 10d = 73 — 38 15d — 10d = 35 5d = 35 d = 7 put common difference (d) value in equation (i) 38 = a + 10×7 38 = a + 70 a = —32 For find a31 Now we have a = —32 and d  = 7 We know that an = a + (n—1)d a31 = a + (31 — 1)d a31 = (—32) + (31 — 1) × 7 a31 = —32 + 30 × 7 a31 = —32 + 210 a31 = 178 So, the 178 is the 31st term of given AP.

Q8.

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Soloution.

Given – a50 = 106 n = 50 a3 = 12 a29 = ? We know that – an = a + (n—1)d For a29 term - Putting value 106 = a + (50—1) d 106 = a + 49d ……………….. (i) Putting value 12 = a + (3—1) d 12 = a + 2d ……………….. (ii) From equation (i) and (ii) (a + 49d) — (a + 2d) = 106 — 12 49d — 2d = 94 47d = 94 d = 2 put common difference (d) value in equation (ii) 12 = a + 2×2 12 = a + 4 a = 8 For find a29 Now we have a = 8 and d  = 2 We know that an = a + (n—1)d a29 = a + (29 — 1)d a29 = 8 + (29 — 1) ×2 a29 = 8 + 28 × 2 a29 = 8 + 56 a29 = 64 So, the 64 is the 29th term of given AP.

Q9.

If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

Solution.

Given – a3 = 4 a9 = —8 an = 0 We know that – an = a + (n—1)d For a3 term - Putting value 4 = a + (3—1) d 4 = a + 2d ……………….. (i) For a9 term - Putting value —8 = a + (9—1) d —8 = a + 8d ……………….. (ii) From equation (i) and (ii) (a + 8d) — (a + 2d) = —8 — 4 a + 8d — a — 2d = —8 — 4 8d — 2d = —12 6d = —12 d = —2 put common difference (d) value in equation (i) 4 = a + 2×(—2) 4 = a — 4 a = 8 For find an =0 Now we have a = 8 and d  = —2 We know that an = a + (n—1)d 0 = 8 + (n — 1)(—2) 0 = 8 + (—2) n — 1(—2) 0 = 8 —2n + 2 0 = 8+2 — 2n 2n = 10 n = 10/2 = 5 So, the 5th term of given AP is Zero.

Q10.

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution.

Givten – The 17th term exceeds the 10th term by 7, this can be written as: a17 — a10 = 7 ,,,,,,,,,,,,,,,,,,,, (i) We know that – an = a + (n—1)d For a17 term - Putting value a17 = a + (17—1) d a17 = a + 17d—d a17 = a + 16d ……………….. (ii) For a9 term - a10 = a + (10—1) d a10 = a + 10d—d a10 = a + 9d ……………….. (iii) Putting value from equation (ii) and (iii) in equation (i) (a + 16d) — (a + 9d) = 7 a + 16d — a — 9d = 7 16d — 9d = 7 7d = 7 d = 1 So, the common difference (d) is 1.

Q11.	Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Solution.	Given – a = 3 d = 15—3 = 12 n = 54 We know that – an = a + (n—1)d For a54 term - Putting value a54 = 3 + (54—1) 12 a54 = 3 + 53(12) a54 = 3 + 636 a54 = 639 As per question, we need to calculate term which is 132 more than 54th term It should be 639 + 132 = 771 Therefore an = 771 n = 1 + (an—a)/d n = 1 + (771—3)/12 n = 1 + 768 / 12 n = 1+ 64 n = 65 so 65th term is 132 more than 54th term.

Q12.

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution.

Given – Common difference for both AP = 100 So we can write – a100—b100 = 100 …………. (i) We know that – an = a + (n—1)d For First AP 100th term a100 = a + (100—1)d a100 = a + 99d …………….. (ii) For Second AP 100th term b100 = b + (100—1)d b100 = b + 99d …………….. (iii) putting value from (ii) and (iii) to (i) (a+99d) —(b+99d) = 100 a + 99d — b — 99d  = 100 a — b = 100 …………… (iv) Now, Difference between 1000th terms- = a1000—b1000 = [a + (1000—1)d]—[ b + (1000—1)d] = (a + 999d)—(b + 999d) = a + 999d—b — 999d = a—b a—b = 100 (by putting value from equation (iv) so difference between 1000th term is also 100.

Q13.

How many three-digit numbers are divisible by 7?

Solution.

We know that series of number divisible by 7 are 7, 14, 21, 28, …. Lowest three digit number divisible by 7 100/7 = 14.28 101/7 = 14.42 102/7 = 14.57 103/7 = 14.71 104/7=14.85 105/7 = 15.00 So the lowest number is 105 Highest three digit number divisible by 7 999/7 = 142.71 998/7 = 142.57 997/7 = 142.42 996/7 = 142.28 995/7=142.14 994/7 = 142 So the highest number is 142 So series will be 105, 112, 119, 126 ….. 994 Here, a = 105 an = 994 d = 7 n (number of term) = ? We know that n = 1 + (an—a)/d n = 1 + (994—105)/7 n = 1 + 889 /7 n = 1+ 127 n = 128 So, 128 three digit number divisible by 7.

Q14.

How many multiples of 4 lie between 10 and 250?

Solution.

We know that series of number multiply of 4 are 4, 8, 12, 16, …. Lowest multiply of  4 between 10 and 252 10/4 = 2.5 11/4 = 2.75 12/4 = 3.00 So the lowest multiply is 12 Highest multiply of  4 between 10 and 252 250/4 = 62.5 249/4 = 62.25 248/4 = 62.00 So the highest multiply is 248 So series will be 12, 16, 20, ….. 248 Here, a = 12 an = 248 d = 4 n (number of term) = ? We know that n = 1 + (an—a)/d n = 1 + (248—12)/4 n = 1 + 236 /4 n = 1+ 59 n = 60 So, 60 multiply of 4 lie between 10 and 250.

Q15.

For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Solution.

We know that – an = a + (n—1)d For 1st AP: Given- a = 63 d = 65—63 = 2 putting value an = 63 + (n—1)2 an =  63 + 2n—2 an =  61 + 2n For 2nd AP: Given- a = 3 d = 10—3 = 7 putting value an = 3 + (n—1)7 an =  3 + 7n—7 an + 7—3 =  7n an + 4 =  7n an =  7n — 4 As per question nth term of first AP = nth term of second AP putting value 61 + 2n = 7n — 4 61 + 4 = 7n —2n 65 = 5n n = 13 So nth term value is 13

Q16.

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution.

We know that an = a, a + d, a + 2d, a + 3d, a + 4d, … As per question third term (a3) a3 = a + 2d = 16 …………….. (i) Fifth term – 7th term (a7) We know that it is equal to a7 = a + 6d ………………. (ii) According to question: a7 — a5 = 12 …………. (ii) putting value (a + 6d) — (a + 6d) = 12 a + 6d — a — 4d = 12 2d = 12 d = 6 put value in equation (i) a + 2×6 = 16 a + 12 = 16 a = 16 —12 a = 4 Because first term is 4 and common difference is 6 so AP is – 4, 10 , 16, 22, 28, …

Q17.	Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
Solution.	Given – AP : 3, 8, 13, . . ., 253 d = 8 —3 = 5 According to question, the 20th term from the last term, so we can consider the last term of the AP as the first term (a1) and work backward and formula changed as per: an = a — (n—1)d a20 = 253 — (20—1)5 a20 = 253 — (19)5 a20 = 253 — 95 a20 = 158 So, the 20th term from the last term of the given AP is 158.

Q18.

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution.

We know that – an = a + (n—1)d and an = a, a + d, a + 2d, a + 3d, a + 4d, … Given – The sum of the 4th and 8th terms is 24: a4 + a8 = 24 a + 3d + a + 7d = 24 2a + 10d = 24 ……………… (i) The sum of the 6th and 10th terms is 44: a6 + a10 = 44 a + 5d + a + 9d = 44 2a + 14d = 44 …………. (ii) Subtract equation (i) from equation (ii) (2a + 14d) —(2a + 10d) =44 — 24 2a + 14d —2a — 10d =44 — 24 2a —2a + 14d — 10d = 44 — 24 14d — 10d = 44 — 24 4d  = 20 d = 5 putting value in equation (i) 2a + 10(5) = 24 2a + 50 = 24 2a  = 24 —50 2a  =  —26 a = —13 So, the first term is —13 and common difference is 5 Second term is = —13 + 5 = —8 Third term is = —8 + 5 = —3 Therefore AP is : —13, —8, —3

Q19.	Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000?
Solution.	Given – d = 200 a = 5000 an = 7000 Salary in 1995  = (a) = 5000 Salary in 1996 = (a+d) 5000 + 200 = 5200 Salary in 1997 = (a+2d) 5000 + 2(200) = 5400 So series became- 5000, 5200, 5400, …. n = ? We know that n = 1 + (an—a)/d n = 1 + (7000—5000)/200 n = 1 + 2000 /200 n = 1+ 10 n = 11 So, Subba rao salary reach Rs. 7000 after 11 year which is 1995 + 11 year = 2006

Q20.	Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs. 20.75, find n.
Solution.	Given – d = 1.75 a = 5 an = 20.75 Saving in 1st week = (a) = 5 Saving in 2nd week = (a+d) 5 + 1.75 = 6.75 Saving in 3rd week  = (a+2d) 5 + 2(1.75) = 8.5 So series became- 5, 6.75, 8.5, …. n = ? We know that n = 1 + (an—a)/d n = 1 + (20.75—5)/1.75 n = 1 + 15.75 /1.75 n = 1+ 9 n = 10 So, in 10 week his saving become Rs. 20.75