Chapter 5: Arithmetic Progressions (NCERT Solutions)

Exercise 5.1: Introduction

Q1. In which of the following situations, does the list of numbers involve make an AP?

(i) Taxi fare: 15, 23, 31, 39... Common diff is 8. Yes.

(ii) Amount of air: Let vol be V. V, 3V/4, 9V/16... Ratio is 3/4. Not AP (It's GP). No.

(iii) Cost of digging: 150, 200, 250... Diff is 50. Yes.

(iv) Compound Interest: 10000(1.08), 10000(1.08)²... Not AP. No.

Q2. Write first four terms...

(i) a=10, d=10: 10, 20, 30, 40.

(ii) a=-2, d=0: -2, -2, -2, -2.

(iii) a=4, d=-3: 4, 1, -2, -5.

(iv) a=-1, d=1/2: -1, -0.5, 0, 0.5.

(v) a=-1.25, d=-0.25: -1.25, -1.50, -1.75, -2.00.

Q3. Write first term and common difference.

(i) 3, 1, -1, -3: a = 3, d = -2.

(ii) -5, -1, 3, 7: a = -5, d = 4.

(iii) 1/3, 5/3, 9/3: a = 1/3, d = 4/3.

(iv) 0.6, 1.7, 2.8: a = 0.6, d = 1.1.

Q4. Which are APs? If yes, find d and next three terms.

(i) 2, 4, 8, 16: Ratio 2. Not AP.

(ii) 2, 5/2, 3, 7/2: d = 0.5. Yes. 4, 9/2, 5.

(iii) -1.2, -3.2, -5.2: d = -2. Yes. -7.2, -9.2, -11.2.

(iv) -10, -6, -2, 2: d = 4. Yes. 6, 10, 14.

(v) 3, 3+√2, 3+2√2: d = √2. Yes.

(vi) 0.2, 0.22, 0.222: Not AP.

(xii) √2, √8, √18: √2, 2√2, 3√2. d = √2. Yes. 4√2, 5√2, 6√2 (√32, √50, √72).

Exercise 5.2: nth Term

Q1. Fill in the blanks.

(i) a=7, d=3, n=8: a_n = 7 + 7(3) = 28.

(ii) a=-18, n=10, a_n=0: 0 = -18 + 9d ⇒ d = 2.

(iii) d=-3, n=18, a_n=-5: -5 = a + 17(-3) ⇒ a = 46.

(iv) a=-18.9, d=2.5, a_n=3.6: 3.6 = -18.9 + (n-1)2.5 ⇒ 22.5 = (n-1)2.5 ⇒ 9 = n-1 ⇒ n = 10.

Q2. Choose correct choice.

(i) 30th term of 10, 7, 4...
a=10, d=-3. a₃₀ = 10 + 29(-3) = 10 - 87 = -77 (C).

(ii) 11th term of -3, -1/2, 2...
d = 2.5 = 5/2. a₁₁ = -3 + 10(5/2) = -3 + 25 = 22 (B).

Q3. Find missing terms.

(i) 2, _, 26: (2+26)/2 = 14.

(ii) _, 13, _, 3: d = (3-13)/2 = -5. Terms: 18, 13, 8, 3.

(iii) 5, _, _, 9.5: 3d = 4.5 ⇒ d = 1.5. Terms: 5, 6.5, 8, 9.5.

Q4. Which term of 3, 8, 13, 18 ... is 78?

78 = 3 + (n-1)5 ⇒ 75 = 5(n-1) ⇒ 15 = n-1 ⇒ n = 16.

Q5. Find number of terms.

(i) 7...205: 205 = 7 + (n-1)6 ⇒ 198/6 = n-1 ⇒ 33 = n-1 ⇒ n = 34.

(ii) 18...-47: -47 = 18 + (n-1)(-2.5) ⇒ -65 = -2.5(n-1) ⇒ 26 = n-1 ⇒ n = 27.

Q6. Is -150 a term of 11, 8, 5, 2...?

-150 = 11 + (n-1)(-3) ⇒ -161 = -3(n-1).
161 is not divisible by 3. No.

Q7. Find 31st term, given 11th is 38 and 16th is 73.

a+10d=38, a+15d=73 ⇒ 5d=35 ⇒ d=7. a=-32.
a₃₁ = -32 + 30(7) = 178.

Q13. How many three digit numbers are divisible by 7?

105, 112... 994.
994 = 105 + (n-1)7 ⇒ 889/7 = n-1 ⇒ 127 = n-1 ⇒ n = 128.

Q16. Determine AP whose 3rd term is 16 and 7th exceeds 5th by 12.

a₇ - a₅ = 12 ⇒ 2d = 12 ⇒ d = 6.
a + 12 = 16 ⇒ a = 4.
AP: 4, 10, 16, 22...

Exercise 5.3: Sum of First n Terms

Q1. Find sum.

(i) 2, 7, 12... to 10 terms: S = 5[4+45] = 245.

(ii) -37, -33... to 12 terms: S = 6[-74+33] = 6(-41)? Wait, d=4. S=6[-74+44] = 6(-30) = -180.

Q2. Find sum (last term given).

(i) 7 + 10.5 + ... + 84: 84 = 7 + (n-1)3.5 ⇒ 77 = 3.5(n-1) ⇒ 22 = n-1 ⇒ n=23.
S = 23/2 [7 + 84] = 23/2(91) = 1046.5.

Q3. In an AP:

(i) a=5, d=3, an=50: n=16, S=440.

(ii) a=7, a13=35: d=7/3, S=273.

(iv) a3=15, S10=125: d=-1, a=17, a10=8.

Q10. Show that a1, a2... form AP where an = 3 + 4n. Find S15.

a1=7, a2=11, a3=15. d=4. Yes AP.
S15 = 15/2 [14 + 14(4)] = 15/2 (70) = 15 × 35 = 525.

Q12. Sum of first 40 positive integers divisible by 6.

6, 12, ... 240.
S = 20 [12 + 234] = 20 × 246 = 4920.

Arithmetic Progressions - RD Sharma Important Questions

Q1. Find p if 2p + 1, 13, 5p - 3 are in AP.

2(13) = (2p + 1) + (5p - 3).
26 = 7p - 2.
28 = 7p ⇒ p = 4.

Q2. If the nth term of a progression is (4n - 10), show that it is an AP. Find its first term and common difference.

a_n = 4n - 10.
a₁ = 4(1) - 10 = -6.
a₂ = 4(2) - 10 = -2.
d = a₂ - a₁ = -2 - (-6) = 4.
Since a_n is linear in n, it is an AP with a = -6, d = 4.

Q3. Determine the AP whose 3rd term is 5 and the 7th term is 9.

a + 2d = 5.
a + 6d = 9.
Subtract: 4d = 4 ⇒ d = 1.
a + 2 = 5 ⇒ a = 3.
AP: 3, 4, 5, 6...

Q4. Find the middle term of the AP: 6, 13, 20, ..., 216.

a = 6, d = 7, l = 216.
216 = 6 + (n-1)7 ⇒ 210 = 7(n-1) ⇒ n = 31.
Middle term = (31+1)/2 = 16th term.
a₁₆ = 6 + 15(7) = 6 + 105 = 111.

Q5. Which term of the AP: 24, 21, 18, ... is the first negative term?

a = 24, d = -3.
a_n < 0 ⇒ 24 + (n-1)(-3) < 0.
24 - 3n + 3 < 0 ⇒ 27 < 3n ⇒ n> 9.
10th term is the first negative term.

Q6. Find the sum of all two digit natural numbers which are divisible by 4.

12, 16, ..., 96.
a = 12, d = 4, l = 96.
96 = 12 + (n-1)4 ⇒ 84 = 4(n-1) ⇒ n = 22.
Sum = 22/2 [12 + 96] = 11 × 108 = 1188.

Q7. If sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find sum of first n terms.

S₇ = 49 ⇒ 7².
S₁₇ = 289 ⇒ 17².
Pattern suggests S_n = n².
Proof: S_n = n/2[2a+(n-1)d]. If a=1, d=2 (odd numbers), sum is n².

Q8. Find the sum of first 20 terms of an AP in which 3rd term is 7 and 7th term is two more than thrice of its 3rd term.

a₃ = 7 ⇒ a + 2d = 7.
a₇ = 3(7) + 2 = 23 ⇒ a + 6d = 23.
4d = 16 ⇒ d = 4.
a + 8 = 7 ⇒ a = -1.
S₂₀ = 20/2 [2(-1) + 19(4)] = 10 [-2 + 76] = 740.

Q9. 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 AP.

(a+3d) + (a+7d) = 24 ⇒ 2a + 10d = 24 ⇒ a + 5d = 12.
(a+5d) + (a+9d) = 44 ⇒ 2a + 14d = 44 ⇒ a + 7d = 22.
Subtract: 2d = 10 ⇒ d = 5.
a + 25 = 12 ⇒ a = -13.
AP: -13, -8, -3.

Q10. How many terms of the AP 24, 21, 18, ... must be taken so that their sum is 78?

78 = n/2 [2(24) + (n-1)(-3)].
156 = n [48 - 3n + 3] = n [51 - 3n].
156 = 51n - 3n².
3n² - 51n + 156 = 0.
n² - 17n + 52 = 0.
(n-13)(n-4) = 0. n = 4 or 13. (Explanation: terms from 5th to 13th likely sum to 0).

Q11. If the sum of first m terms of an AP is equal to sum of first n terms, then show that the sum of its (m+n) terms is zero.

S_m = S_n.
m/2 [2a + (m-1)d] = n/2 [2a + (n-1)d].
2a(m-n) + d(m(m-1) - n(n-1)) = 0.
2a(m-n) + d((m-n)(m+n) - (m-n)) = 0.
Div by (m-n): 2a + d(m+n-1) = 0.
S_m+n = (m+n)/2 [2a + (m+n-1)d] = (m+n)/2 [0] = 0.

Q12. The sum of the first n terms of an AP is given by S_n = 3n² - 4n. Find the nth term.

a_n = S_n - S_n-1.
(3n² - 4n) - [3(n-1)² - 4(n-1)].
(3n² - 4n) - [3(n²-2n+1) - 4n+4].
3n² - 4n - 3n² + 6n - 3 + 4n - 4.
6n - 7.

Q13. If m times the mth term of an AP is equal to n times the nth term, prove that its (m+n)th term is 0.

m[a + (m-1)d] = n[a + (n-1)d].
a(m-n) + d[m²-m - n²+n] = 0.
a(m-n) + d[(m-n)(m+n) - (m-n)] = 0.
Div by (m-n): a + d(m+n-1) = 0.
This is exactly a_m+n. So 0.

Q14. Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.

Let parts be a-d, a, a+d.
Sum = 3a = 207 ⇒ a = 69.
(69-d)(69) = 4623.
69-d = 67 ⇒ d = 2.
Parts: 67, 69, 71.

Q15. The sum of the first n terms of an AP is 3n² + 6n. Find the nth term and the 15th term.

a_n = S_n - S_n-1 = 6n + 3.
a₁₅ = 6(15) + 3 = 93.

Q16. Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.

S₄ = 1/2 (S₈ - S₄) ⇒ 2S₄ = S₈ - S₄ ⇒ 3S₄ = S₈.
3 [4/2 (2a + 3d)] = 8/2 (2a + 7d).
6 (10 + 3d) = 4 (10 + 7d).
60 + 18d = 40 + 28d.
20 = 10d ⇒ d = 2.

Q17. Solve the equation: 1 + 4 + 7 + 10 + ... + x = 287.

a = 1, d = 3. Let there be n terms.
n/2 [2 + (n-1)3] = 287.
n(3n - 1) = 574.
3n² - n - 574 = 0.
n = 14.
x = a₁₄ = 1 + 13(3) = 40.

Q18. The sum of n terms of two APs are in ratio (5n+4):(9n+6). Find ratio of their 18th terms.

Ratio of sums S_n/S'_n = (2a+(n-1)d) / (2a'+(n-1)d').
To get ratio of 18th terms (a+17d)/(a'+17d'), put (n-1)/2 = 17 ⇒ n = 35.
Ratio = (5(35)+4)/(9(35)+6) = 179/321.

Q19. If the pth term of an AP is 1/q and the qth term is 1/p, prove that the sum of first pq terms is 1/2(pq + 1).

Standard result. a = 1/pq, d = 1/pq.
S_pq = pq/2 [2/pq + (pq-1)/pq] = pq/2 [(2+pq-1)/pq] = 1/2(pq+1).

Q20. Find sum of all integers between 100 and 550 which are not divisible by 9.

Total sum - Sum of divisible by 9.
Total: 101 to 549. n = 449. S = 449/2(650) = 145925.
Div by 9: 108, ..., 549. n = 50. S = 50/2(108+549) = 16425.
Req Sum = 145925 - 16425 = 129500.

Arithmetic Progressions - Formulas & PYQs

Key Formulas & Concepts

1. General Term (nth Term)

a_n = a + (n - 1)d

where a is first term, d is common difference.

2. Sum of First n Terms

S_n = n/2 [2a + (n - 1)d]

S_n = n/2 (a + l)

where l is the last term.

3. Sum of First n Positive Integers

S_n = n(n + 1) / 2

Previous Year Questions (CBSE/JKBOSE)

Q1. Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253. (CBSE 2017)

Reverse AP: 253, 248, ...
a = 253, d = -5.
a₂₀ = 253 + 19(-5) = 253 - 95 = 158.

Q2. If the sum of first m terms of an AP is 2m² + 3m, then what is its second term? (CBSE 2016)

S₁ = a₁ = 2 + 3 = 5.
S₂ = 2(4) + 6 = 14.
a₂ = S₂ - S₁ = 14 - 5 = 9.

Q3. Which term of the AP 21, 18, 15, ... is zero? (CBSE 2015)

a = 21, d = -3.
0 = 21 + (n - 1)(-3).
3(n - 1) = 21 ⇒ n - 1 = 7 ⇒ n = 8.

Q4. Find standard deviation? (Out of syllabus for this chapter usually, checking context). Correction: Question meant Arithmetic Mean usually. Q4. Find sum of odd numbers between 0 and 50. (CBSE 2015)

1, 3, ..., 49.
n = 25.
Sum = n² = 25² = 625.

Q5. Determine k so that k+2, 4k-6 and 3k-2 are three consecutive terms of an AP. (CBSE 2019)

2(4k-6) = (k+2) + (3k-2).
8k - 12 = 4k.
4k = 12 ⇒ k = 3.

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