# Chapter 5 Arithmetic Progressions Important Questions for CBSE Class 10 Maths Board Exams

**Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions**, which will help the students to prepare for the CBSE Class 10 maths Board exam 2022-23. It help the doing better in their maths paper. Extra questions of Chapter 5 Arithmetic Progressions given here which are based on the pattern of CBSE NCERT book. Students will learn about the entire syllabus and learn how to solve problems in preparation for the exams.

## Important Questions for Chapter 5 Arithmetic Progressions Class 10 Maths

### Arithmetic Progressions Class 10 Maths Important Questions Very Short Answer (1 Mark)

**1. The sum of first 20 terms of the AP 1, 4, 7, 10 .... is**

**Solution**

**2. If 4 times the 4th term of an AP is equal to 18 times the 18th term, then find the 22 nd term.**

**Solution**

_{n}= a + (n-1)d

_{4}= 18 × a

_{18}

_{22}= 0

**3. Find the sum of all 11 terms of an AP whose middle term is 30.**

**Solution**

_{6}= a + 5d = 30

**4. If the first three terms of an AP are b, c and 2b, then find the ratio of b and c.**

**Solution**

**5. Find the 9th term from the end (towards the first term) of the A.P. 5,9,13, …, 185.**

**Solution**

Here,

First term, a = 5

Common difference, d = 9 – 5 = 4

Last term, 1 = 185

n^{th} term from the end = l – (n – 1)d

9^{th} term from the end = 185 – (9 – 1)4

= 185 – 8 × 4

= 185 – 32

= 153

**6. Find the common difference of the AP 1/p, 1−p/p, 1−2p/p,…**

**Solution**

The common difference,

**7. What is the common difference of an A.P. in which a**

_{21}– a_{7}= 84?**Solution**

_{21}– a

_{7}= 84

**…[Given]**

**…[a**

_{n }= a + (n – 1)d]**8. Find, how many two digit natural numbers are divisible by 7.**

**Solution**

_{n}= 98

_{n }= a + (n – 1)d

**9. Which of the term of AP 5, 2, ,...... 1 - is 49 - ?**

**Solution**

_{n }= a + (n – 1)d

**10. If the common difference of an AP is -6, find a**

_{16}- a_{12}.**Solution**

_{16}= a + (16-1)(-6) = a - 90

_{12}= a + (12-1)(-6) = a - 66

_{16}- a

_{12}= (a-90) - (a-66)

### Arithmetic Progressions Class 10 Maths Important Questions Short Answer-I(2 Marks)

**11. Which term of the progression 4, 9, 14, 19, … is 109? (2011D)**

**Solution**

Here,

d = 9 - 4 = 14 -9 = 19 – 14 = 5

∴ Difference between consecutive terms is constant.

Hence, it is an A.P.

Given,

First term, a = 4, d = 5, a_{n} = 109 **(Let)**

∴ a_{n} = a + (n – 1) d **…[General term of A.P.]**

∴ 109 = 4 + (n – 1) 5

⇒ 109 – 4 = (n – 1) 5

⇒ 105 = 5(n − 1)

⇒ n – 1 = 105/5 = 21

⇒ n = 21 + 1 = 22

∴ 109 is the 22^{nd} term

**12. Find the sum of first ten multiple of 5.**

**Solution**

Let the first term be a, common difference be d, nth term be a_{n} and sum of n term be S_{n}.

Here,

a = 5, n = 10, d = 5

= 5[10+ 9×5]

= 5[10+45]

= 5×55 = 275

Hence, the sum of first ten multiple of 5 is 275.

**13. If the sum of first m terms of an AP is the same as the sum of its first n terms, show that the sum of its first (m+n) terms is zero.**

**Solution**

Let a be the first term and d be the common difference of the given AP.

Then,

S_{m} = S_{n}

**14. If 3k-2, 4k-6 and k+2 are three consecutive terms of AP, then find the value of k.**

**Solution**

**15. How many terms of AP 3, 5, 7, 9, ..... must be taken to get the sum 120?**

**Solution**

_{n}= 120

^{2}+ 2n - 120 = 0

^{2}+ 12n - 10n - 120 = 0

**16. Which term of the AP 3, 15, 27, 39, ... will be 120 more than its 21st term?**

**Solution**

_{n}= a + (n-1)d

_{21}= 3 + (21-1) 12

_{n}= a + (n – 1) d

**17. Find the 21st term of the AP -4 1/2, -3, -1 1/2, ....**

**Solution**

**18. The 4**

^{th}term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11^{th}term.**Solution**

Let 1^{st} term = a, Common difference = d

a_{4} = 0 a + 3d = 0

⇒ a = -3d **…(i)**

To prove: a_{25} = 3 × a_{11}

a + 24d = 3(a + 10d) **…[From (i)]**

⇒ -3d + 24d = 3(-3d + 10d)

⇒ 21d = 21d

From above, a_{25} = 3(a_{11})

**19. Find 10th term from end of the A.P. 4,9, 14, …, 254. (2011OD)**

**Solution**

Common difference d = 9 – 4

= 14 – 9 = 5

Given: Last term, l = 254, n = 10

n^{th} term from the end = l – (n – 1) d

∴ 10^{th} term from the end = 254 – (10 – 1) × 5

= 254 – 45 = 209

**20. Find, 100 is a term of the AP ****25, 28, 31****,.... or not.**

**Solution**

Let the first term of an AP be a, common difference

be d and number of terms be n.

Let a_{n} = 100

Here, a = 25, d = 28-25 = 31-28 = 3

Now,

a_{n} = a + (n-1)d,

100 = 25 + (n-1) 3

⇒ 100 - 25 = 75 = (n-1) 3

⇒ 25 = n - 1

⇒ n = 26

Since, 26 is an whole number, thus 100 is a term of given A.P.

**21. Find the 7th term from the end of AP 7, 10, 13,.... 184**

**Solution**

Let us write AP in reverse order i.e., 184, .....13, 10, 7

Let the first term of an AP be a and common

difference be d.

Now,

d = 7-10 = -3

a = 184, n = 7

7th term from the original end.

a_{7} = a +6d

⇒ a_{7} = 184 + 6(-3)

= 184 -18 = 166

Hence, 166 is the 7th term from the end.

**22. Which term of an AP 150, 147, 144, ..... is its first negative term?**

**Solution**

_{n}.

_{n}< 0

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

**Solution**

_{17}= a

_{10}+ 7

**24. The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.**

**Solution**

Here, a_{n} = 45, S_{n} = 400, a = 5, n = ?, d = ?

**25. How many terms of the AP 65, 60, 55.... 65 60 55 be taken so that their sum is zero?**

**Solution**

Let the first term be a, common difference be d, nth term be a n and sum of n term be Sn .

We have, a = 65, d = -5, S_{n} = 0

⇒ 135n - 5n^{2} = 0

⇒ n(135 - 5n) = 0

⇒ 5n = 135

⇒ n = 27

**26. How many terms of the AP 18, 16, 14.... be taken so that their sum is zero?**

**Solution**

Let the first term be a, common difference be d, nth term be an and sum of n term be S_{n}.

Here,

a = 18, d = -2, S_{n} = 0

**27. The first and last term of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.**

**Solution**

_{n}.

_{n}= 400

**28. Find the sum of all three digit natural numbers, which are multiples of 11.**

**Solution**

To find: 110 + 121 + 132 + … + 990

Here,

a = 110, d = 121- 110 = 11, a_{n} = 990

∴ a + (n – 1)d = 990

110 + (n – 1).11 = 990

⇒ (n – 1). 11 = 990 – 110 = 880

⇒ (n – 1) = 880 = 80

⇒ n = 80 + 1 = 81

As,

S_{n} = n/2 (a_{1} + a_{n})

∴ S_{81} = 81/2 (110 + 990)

= 81/2 (1100) = 81 × 550 = 44,550

**29. Find how many integers between 200 and 500 are divisible by 8.**

**Solution**

Number divisible by 8 are 208, 2016, 224, .... 496. It is an AP

Let the first term be a, common difference be d and nth term be a_{n}.

We have a = 208, d = 8, a_{n} = 496

Now,

a + (n-1)d = a_{n}

⇒ 208 + (n-1)d = 496

⇒ (n-1)8 = 496 - 208

⇒ n = 36+1 = 37

Hence, required numbers divisible by 8 is 37.

**30. The fifth term of an AP is 26 and its 10th term is 51. Find the AP**

**Solution**

Let the first term be a, common difference be d and nth term be a_{n}.

a_{5} = a + 4d = 26 **...(1)**

a_{10} = a + 9d = 51 **...(2)**

Subtracting (1) from (2) we have

5d = 25

⇒ d = 5

Substituting this value of d in equation (1) we get

a = 6

Hence, the AP is 6, 11, 16, ....

**32. Find whether -150 is a term of the AP 11, 8, 5, 2, ...**

**Solution**

Let the first term be a, common difference be d and nth term be a_{n}.

Let the nth term of given AP 11, 8, 5, 2... be -150

Hence,

a = 11, d = 8-11 = -3 and an = -150

a + (n-1)d = an

⇒ 11 + (n-1)(-3) = -150

⇒ (n-1)(-3) = -161

which is not a whole number. Hence -150 is not a term of given AP.

### Arithmetic Progressions Class 10 Maths Important Questions Short Answer-II (3 Marks)

**33. The sum of the first 7 terms of an AP is 63 and that of its next 7 terms is 161. Find the AP.**

**Solution**

_{7}= 63

**34. The sum of four consecutive number in AP is 32 and the ratio of the product of the first and last term to the product of two middle terms is 7 : 15. Find the numbers.**

**Solution**

**35. Which term of the A.P. 3, 14, 25, 36,… will be 99 more than its 25th term?**

**Solution**

Let the required term be n^{th} term, i.e., a_{n}

Here,

d = 14 – 3 = 11, a = 3

According to the Question,

a_{n} = 99 + a_{25}

∴ a + (n – 1) d = 99 + a + 24d

⇒ (n – 1) (11) = 99 + 24 (11)

⇒ (n – 1) (11) = 11 (9 + 24)

⇒ n – 1 = 33

⇒ n = 33 + 1 = 34

∴ 34^{th} term is 99 more than its 25th term.

**36. Which term of the AP 20, 19 1/4, 18 1/2, 17 3/4, ...is the first negative term.**

**Solution**

Here,

a = 20

n = 28

Hence, the first negative term is 28th term.

**37. If in an AP, the sum of first m terms is n and the sum of its first n terms is m, then prove that the sum of its first (m+n) terms is -(m+n)**

**Solution**

Let 1st term of series be a and common difference be d, then we have

S_{m} = n

S_{n} = m

**38. Find the 20th term of an AP whose 3rd term is 7 and the seventh term exceeds three times the 3rd term by 2. Also find its nth term (a**

_{n}).**Solution**

_{3}= a+2d = 7

**...(1)**

_{7}= 3a3 + 2

**...(2)**

_{20}= a+19d = -1 + 19×4 = 75

_{n}= a+(n-1)d

**39. If 7th term of an AP is 1/9 and 9th term is 1/7, find 63rd term.**

**Solution**

_{n}.

**40. The 19th term of an AP is equal to three times its 6**

^{th}term. If its 9^{th}term is 19, find the A.P.**Solution**

Given,

a_{19} = 3(a_{6})

⇒ a + 18d = 3(a + 5d)

⇒ a + 18d = 3a + 15d

⇒ 18d – 15d = 3a – a

**41. The sum of the 5th and the 9th terms of an AP is 30. If its 25**

^{th}term is three times its 8^{th}term, find the AP.**Solution**

a_{5} + a_{9} = 30 **…[Given]**

⇒ a + 4d + a + 8d = 30 **…[∵a _{n} = a + (n – 1)d]**

⇒ 2a + 12d = 30

⇒ a + 6d = 15

**…[Dividing by 2]**

⇒ a = 15 – 6d

**…(i)**

Now, a

_{52}= 3(a

_{8})

a + 24d = 3(a + 7d)

⇒ 15 – 6d + 240 = 3(15 – 6d + 7d)

**…[From (i)]**

⇒ 15 + 18d = 3(15 + d)

⇒ 15 + 18d = 45 + 3d

⇒ 18d – 3d = 45 – 15

⇒ 15d = 30

∴ d = 30/15 = 2

From (i), a = 15 – 6(2) = 15 – 12 = 3

**42. The ninth term of an AP is equal to seven times the second term and twelfth term exceeds five times the third term by 2. Find the first term and the common difference.**

**Solution**

Let the first term be a, common difference be d and nth term be a_{n}.

Now,

a_{9} = 7a_{2}

⇒ a + 8d = 7(a+d)

⇒ a + 8d = 7a + 7d

⇒ -6a + d = 0 **...(1)**

and

a_{12} = 5a_{3} + 2

⇒ a + 11d = 5(a+2d) + 2

⇒ a + 11d = 5a + 10d + 2

⇒ -4a + d = 2 **...(2)**

Subtracting (2) from (1), we get

-2a = -2

⇒ a = 1

Substituting this value of a in equation (1) we get

-6 + d = 0

⇒ d = 6

Hence first term is 1 and common difference is 6.

**43. The sum of first n terms of three arithmetic progressions are S**

_{1}, S_{2}and S_{3}respectively. The first term of each AP is 1 and common differences are 1, 2 and 3 respectively. Prove that S_{1}+ S_{3}= 2S_{2}.**Solution**

_{1}= 1+2+3+ ... n

_{2}= 1+3+5+ ... up to n terms

_{3}= 1+4+7+ ... upto n terms

**44. If Sn denotes, the sum of the first n terms of an AP prove that S**

_{12}= 3(S_{8}- S_{4}).**Solution**

_{n}.

**45. Find the number of terms of the AP 18, 15 1/2, 13,…, -49 1/2 and find the sum of all its terms.**

**Solution**

Here 1^{st} term, a = 18

**46. If the ratio of the sums of first n terms of two AP’s is (7n + 1): (4n + 27), find the ratio of their nth terms.**

**Solution**

Let a and A be the first term and d and D be the common difference of two AP’s, then we have

**47. The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference of the A.P.**

**Solution**

First term, a = 5, Last term, a_{n} = 45

Let the number of terms = n

S_{n} = 400

⇒ n/2 (a + a_{n}) = 400

⇒ n/2(5 + 45) = 400

⇒ n/2 (50) = 400

⇒ n = 400/25 = 16 = Number of terms

Now, a_{n} = 45

a + (n – 1)d = a_{n}

⇒ 5+ (16 – 1)d = 45

⇒ 15d = 45 – 5

∴ d = 40/15=8/3

**48. The sum of first n terms of an AP is 3n ^{2} + 4n. Find the 25^{th} term of this AP.**

**Solution**

We have, S_{n} = 3n^{2} + 4n

Put n = 25,

S_{25} = 3(25)^{2} + 4(25)

= 3(625) + 100

= 1875 + 100 = 1975

Put n = 24,

S_{24} = 3(24)^{2} + 4(24)

= 3(576) + 96

= 1728 + 96 = 1824

∴ 25^{th} term = S_{25} – S_{24}

= 1975 – 1824 = 151

**49. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.**

**Solution**

Let hundred’s place digit = (a – d)

Let ten’s place digit = a

Let unit’s place digit = a + d

According to the Question,

a – d + a + a + d = 15

⇒ 3a = 15

⇒ a = 5

Original number

= 100(a – d) + 10(a) + 1(a + d)

= 100a – 100d + 10a + a + d

= 111a – 99d

Reversed number

= 1(a – d) + 10a + 100(a + d)

= a – d + 10a + 100a + 100d

= 111a + 99d

Now, Original no. – Reversed no. = 594

111a – 99d – (111a + 99d) = 594

-198d = 594

⇒ d = 594/-198 = -3

∴ The Original no. = 111a – 99d

= 111(5) – 99(-3)

= 555 + 297 = 852

**50. Find the sum of all multiples of 7 lying between 500 and 900. (2012OD)**

**Solution**

To find: 504 + 511 + 518 + … + 896

a = 504, d = 511- 504 = 7, a_{n} = 896

a + (n – 1)d = a_{n}

∴ 504 + (n – 1)7 = 896

(n – 1)7 = 896 – 504 = 392

**51. The tenth term of an AP, is -37 and the sum of its first six terms is -27. Find the sum of its first eight terms.**

**Solution**

_{n}= a + (n-1)d

**...(1)**

**...(2)**

_{n}= -76

**52. The 16th term of an AP is five times its third term. If its 10th term is 41, then find the sum of its first fifteen terms.**

**Solution**

_{16}= 5a

_{3}

**...(1)**

**...(2)**

_{15}= 495

**53. Find the sum of n terms of the series**

**(4 - 1/n) + (4 - 2/n) + (4 - 3/n) +....**

**Solution**

_{n}

**54. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.**

**Solution**

_{n}.

_{n}= 990, d = 10

_{n}= a + (n - 1)d

**55. If the ratio of the 11th term of an AP to its 18th term is 2 : 3, find the ratio of the sum of the first five term of the sum of its first 10 terms.**

**Solution**

_{n}

_{5}: S

_{10}= 6:17

### Arithmetic Progressions Class 10 Maths Important Questions Long Answer (4 Marks)

**56. The first term of an AP is 3, the last term is 83 and the sum of all its terms is 903. Find the number of terms and the common difference of the AP.**

**Solution**

_{n}= 83

_{n}= 903

**57. Find the common difference of the Arithmetic Progression 1/a, 3-a/3a, 3-2a/3a, ... (a≠0)**

**Solution**

**58. f p**

^{th}, q^{th}and r^{th}terms of an A.P. are a, b, c respectively, then show that (a – b)r + (b – c)p+ (c – a)q = 0.**Solution**

Let A be the first term and D be the common difference of the given A.P.

p^{th} term = A + (p – 1)D = a **…(i)**

q^{th} term = A + (q – 1)D = b **…(ii)**

r^{th} term = A + (r – 1)D = c **…(iii)**

L.H.S. = (a – b)r + (b – c)p + (c – a)q

= [A + (p – 1)D – (A + (q – 1)D)]r + [A + (q – 1)D – (A + (r – 1)D)]p + [A + (r – 1)D – (A + (p – 1)D)]q

= [(p – 1 – q + 1)D]r + [(q – 1 – r + 1)D]p + [(r – 1 – p + 1)D]q

= D[(p – q)r + (q – r)p + (r – p)q]

= D[pr – qr + qp – rp + rq – pq]

= D[0] = 0 = R.H.S.

**59. A sum of ₹1,600 is to be used to give ten cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.**

**Solution**

Here,

S_{10} = 1600, d = -20, n = 10

S_{n} = n/2 (2a + (n – 1)d]

∴ 10/2[2a + (10 – 1)(-20)] = 1600

2a – 180 = 320

2a = 320 + 180 = 500

a = 250

∴ 1^{st} prize = a = ₹250

2^{nd} prize = a_{2} = a + d = 250 + (-20) = ₹230

3^{rd} prize = a_{3} = a_{2} + d = 230 – 20 = ₹210

4^{th} prize = a_{4} = a_{3} + d = 210 – 20 = ₹190

5^{th} prize = a_{5} = a_{4} + d = 190 – 20 = ₹170

6^{th} prize = a_{6} = a_{5} + d = 170 – 20 = ₹150

7^{th} prize = a_{7} = a_{6} + d = 150 – 20 = ₹130

8^{th} prize = a_{8} = a_{7} + d = 130 – 20 = ₹110

9^{th} prize = a_{9} = a_{8} + d = 110 – 20 = ₹590

10^{th} prize = a_{10} = a_{9} + d = 90 – 20 = ₹70

= ₹1,600

**60. If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.**

**Solution**

Let a be the first term and d be the common difference.

Sum of n terms of an AP,

Hence, sum of n terms is 6n + n^{2}.

**61. If Sn denotes the sum of first n terms of an AP, prove that, S _{30} = 3(S_{20} - S_{10})**

**Solution**

Let the first term be a, and common difference be d.

Hence, S_{30} = 3(S_{20} - S_{10})

**62. Find the 60th term of the AP 8, 10, 12,.... if it has a total of 60 terms and hence find the sum of its last 10 terms.**

**Solution**

Let the first term be a, common difference be d, nth term be a n and sum of n term be S_{n}

We have,

a = 8, d = 10 - 8 = 2

a_{n} = a + (n-1)d

Now,

a_{60} = 8 + (60 - 1)2

= 8+ 59×2 = 126

and a_{51} = 8 + 50×2

= 8 + 100 = 108

Sum of last 10 terms,

Hence sum of last 10 terms is 1170.

**63. An AP consists of 37 terms. The sum of the three ****middle most terms is 225 and the sum of the past ****three terms is 429. Find the AP.**

**Solution**

Let the middle most terms of the AP be (x-d), x and (x+d).

We have,

x-d + x + x+d = 225

⇒ 3x = 225

⇒ x = 75

and the middle term = 37+1/2 = 19th term.

Thus AP is,

(x-18d),...(x-2d), (x-d), x, (x+d), (x+2d),...(x+18d)

Sum of last three terms,

(x+18d) + (x+17d) + (x+16d) = 429

⇒ 3x + 51d = 429

⇒ 225 + 51d = 429

⇒ d = 4

First term, a_{1} = x-18d = 75- 18×4 = 3

a_{2} = 3+4 = 7

Hence, AP = 3, 7, 11,...,147.

**64. The first and the last terms of an A.P. are 8 and 350 respectively. If its common difference is 9, how many terms are there and what is their sum?**

**Solution**

Here,

a = 8, a_{n} = 350, d = 9

As we know, a + (n − 1) d = a_{2}

∴ 8 + (n – 1) 9 = 350

⇒ (n − 1) 9 = 350 – 8 = 342

⇒ n – 1 = 342/9 = 38

⇒ n = 38 + 1 = 39

∴ There are 39 terms.

∴ S_{n} = n/2(a + an)

∴ S_{39} = 39/2 (8 + 350) = 39/2 × 358

= 39 × 179 = 6981

**65. In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP.**

**Solution**

Here,

n = 50,

S_{10} = 210

= 10/2 (2a + 9d) = 210 **…[S _{n} = 1/2 [2a+(n – 1)2]**

⇒ 5(2a + 9d) = 210

⇒ 2a + 9d = 210/5 = 42

⇒ 2a = 42 – 9d

⇒ a = 42−9d/2

**…(i)**

Now,

50 = (1 + 2 + 3 + …) + (36 + 37 + … + 50) Sum = 2565

Sum of its last 15 terms = 2565

**…(Given)**

S

_{50}– S

_{35}= 2565

⇒ 50/2(2a + 49d) – 35/2 (2a + 34d) = 2565

⇒ 100a + 2450d – 70a – 1190d = 2565 × 2

⇒ 30a + 1260d = 5130

⇒ 3a + 1260 = 513

**…(Dividing both sides by 10)**

**66. In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two Sections, find how many trees were planted by the students.**

**Solution**

Classes: 1 + I + II + … + XII

Sections: 2(I) + 2(II) + 2(III) + … + 2(XII)

Total no. of trees

= 2 + 4 + 6 … + 24

= (2 × 2) + (2 × 4) + (2 × 6) + … + (2 × 24)

= 4 + 8 + 12 + … + 48

:: S_{12} = 12/2(4 + 48) = 6(52) = 312 trees

**67. Find the sum of first 24 terms of an AP whose nth ****term is given by an = 3+2n**

**Solution**

Let the first term be a, common difference be d, nth term be an and sum of n term be Sn

We have,

a_{n} = 3+2n

a_{1} = 3+ 2×1 = 5

a_{2} = 3 + 2×2 = 7

a_{3} = 3 + 2×3 = 9

Thus, the series is 5, 7, 9,... in which

a = 5 and d = 2

Now,

Hence, S_{24} = 672

**68. Ramkali required ₹500 after 12 weeks to send her daughter to school. She saved ₹100 in the first week and increased her weekly saving by ₹20 every week. Find whether she will be able to send her daughter to school after 12 weeks.**

**Solution**

Money required by Ramkali for admission of her daughter = ₹2500

A.P. formed by saving

100, 120, 140, … upto 12 terms **…(i)**

Let, a, d and n be the first term, common difference and number of terms respectively.

Here,

a = 100, d = 20, n = 12

S_{n} = n/2 (2a + (n − 1)d)

⇒ S_{12} = 12/2 (2(100) + (12 – 1)20)

⇒ S_{12} = 12/2 [2(100) + 11(20)] = 6[420] = ₹2520

**69. How many multiples of 4 lie between 10 and 250? Also find their sum. (2011D)**

**Solution**

Multiples of 4 between 10 and 250 are:

12, 16, 20, … 248

Here, a = 12, d = 4, a_{n} = 248

As we know, a + (n – 1) d = a_{n}

∴12 + (n – 1) 4 = 248

⇒ (n – 1) 4 = 248 – 12 = 236

⇒ n – 1 = 236/4 = 59

⇒ n = 59 + 1 = 60

∴ There are 60 terms.

Now, S_{n} = n2 (a + a_{n})

∴ S_{60} = 60/2(12 + 248)

= 30 (260) = 7800

**70. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.**

**Solution**

Here,

A.P. is 1, 2, 3, …., 49

a = 1, d = 1, a_{n} = 49

Now,

⇒ (X – 1)^{2} (2 + (X – 2)) = 49(2 + 48) – X[2 + (x – 1)]

⇒ (X – 1). X = 2,450 – X(X + 1)

⇒ x^{2} – X = 2,450 – X^{2} – X

⇒ X^{2} – X + X^{2} + X = 2,450

⇒ 2X^{2} = 2,450

⇒ X^{2} = 1,225

∴ X = √1225 = 35 **…[X can not be -ve]**