

InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
501. |
Find(i) 10th term of A.P. 2, 7, 12, ….(ii) 18th term of A.P. √2, 3√2, 5√2, …..(iii) 24th term of A.P. 9, 13, 17, 21, …. |
Answer» (i) Given A.P. 2, 7, 12, ….. First term a = 2 Common difference d = 7 – 2 = 5 nth term an = a + (n – 1)d ∴ a10 = 2 + (10 – 1) × 5 = 2 + 9 × 5 = 2 + 45 = 47 Hence, 10th term of given series 47. (ii) Given A.P. √2, 3√2, 5√2, ….. First term a = √2 Common term d = 3√2 – √2 = √2(3 – 1) = 2√2 nth term an = a + (n – 1)d ∴ a18 = √2 + (18 – 1)2√2 = √2 + 17 × 2√2 = √2 + 34√2 = 35√2 Hence a18 = 35√2 (iii) Given A.P. 9, 13, 17, 21, ….. First term a = 9 common difference d = 13 – 9 = 4 nth term an = a + (n – 1)d ∴ a24 = 9 + (24 – 1) × 4 = 9 + 23 × 4 = 9 + 92 = 101 Hence a24 = 101 |
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502. |
The nth term of an A.P is given by (-4n + 15). Find the sum of first 20 terms of this A.P. |
Answer» Given, The nth term of the A.P = (-4n + 15) So, by putting n = 1 and n = 20 we can find the first ans 20th term of the A.P a = (-4(1) + 15) = 11 And, a20 = (-4(20) + 15) = -65 Now, for find the sum of 20 terms of this A.P we have the first and last term. So, using the formula Sn = \(\frac{n}{2}\)(a + l) S20 = \(\frac{20}{2}\)(11 + (-65)) = 10(-54) = -540 Therefore, the sum of first 20 terms of this A.P. is -540. |
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503. |
Find the 15th term of the AP -40, -15, 10, 35, …… |
Answer» The 15th term of the AP -40, -15, 10, 35, …… Given: AP is -40, -15, 10, 35, …… Here, first term = a = -40 Common difference = d = -15 – (-40) = -15 + 40 = 25 n = 15 an = a + (n-1)d a15 = -40 + (15 – 1)25 = 310 |
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504. |
In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term. |
Answer» Given, 24th term is twice the 10th term. We know that, nth term an = a + (n – 1)d ⇒ a24 = 2(a10) a + (24 – 1)d = 2(a + (10 – 1)d) a + 23d = 2(a + 9d) a + 23d = 2a + 18d a = 5d …. (1) Now, the 72nd term can be expressed as a72 = a + (72 – 1)d = a + 71d = a + 5d + 66d = a + a + 66d [using (1)] = 2(a + 33d) = 2(a + (34 – 1)d) = 2(a34) ⇒ a72 = 2(a34) Hence, the 72nd term is twice the 34th term of the given A.P. |
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505. |
Find:10th term of the AP -40, -15, 10, 35, …………. |
Answer» Given A. P. is – 40, -15, 10, 35, ………. First term (a) = -40 Common difference (d) = Second term – fast term = -15 – (- 40) = 40 – 15 = 25 We know that, nth term of an A.P. an = a + (n – 1)d Then, 10th term of A. P. a10 = -40 + (10 – 1)25 = – 40 + 9.25 = – 40 + 225 = 185 ∴ 10th term of the A. P. is 185. |
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506. |
Find n if the given value of x is the nth term of the given A.P.1, \(\frac{21}{11}\), \(\frac{31}{11}\), \(\frac{41}{11}\), …; x = \(\frac{171}{11}\) |
Answer» Given A.P. 1, \(\frac{21}{11}\), \(\frac{31}{11}\), \(\frac{41}{11}\), \(\frac{171}{11}\) Here, a = 1 d = \(\frac{21}{11}\) – 1 = \(\frac{10}{11}\) Last term (nth term) = \(\frac{171}{11}\) We know that an = a + (n – 1)d ⇒ \(\frac{171}{11}\) = 1 + (n-1)\(\frac{10}{11}\) 171 = 11 + 10n – 10 n = 170/10 n = 17 |
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507. |
If the 10th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth term. |
Answer» Let’s consider a to be the first term and d be the common difference. And we know that, sum of first n terms is: Sn = \(\frac{n}{2}\)(2a + (n − 1)d) and nth term is given by: an = a + (n – 1)d Now, from the question we have S10 = 120 ⟹ 120 = \(\frac{10}{2}\)(2a + (10 − 1)d) ⟹ 120 = 5(2a + 9d) ⟹ 24 = 2a + 9d …. (1) Also given that, a10 = 21 ⟹ 21 = a + (10 – 1)d ⟹ 21 = a + 9d …. (2) Subtracting (2) from (1), we get 24 – 21 = 2a + 9d – a – 9d ⟹a = 3 Now, on putting a = 3 in equation (2), we get 3 + 9d = 21 9d = 18 d = 2 Thus, we have the first term(a) = 3 and the common difference(d) = 2 Therefore, the nth term is given by an = a + (n – 1)d = 3 + (n – 1)2 = 3 + 2n -2 = 2n + 1 Hence, the nth term of the A.P is (an) = 2n + 1. |
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508. |
The 17th term of an AP exceeds its 17th term by 7. Find the common difference. |
Answer» a17 = a10 + 7, d =? a + 16d = a + 9d + 7 a + 16d – a – 9d = 7 7d = 7 ∴ d = \(\frac{7}{7}\) ∴ d = 1 |
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509. |
The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P. |
Answer» S7 = 63 \(\frac{7}{2}\)[2(a) + 6d] = 639 a + 3d = 9 (i) a8 = a + 7d Hence, for next 7 terms first term will be the 8th term i.e. a + 7d Sum of next 7 terms, S’7 = \(\frac{7}{2}\)[2(a + 7d) + 6d] 161 = 7 [a + 7d + 3d] 23 = a + 10d 23 = 9 – 3d + 10d [From (i)] 14 = 7d d = 2 Putting the value of d in (i), we get A = 9 – 3(2) = 3 Now, a28= a + 27d = 3 + 27(2) = 3 + 54 = 57 |
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510. |
The sum of first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P. |
Answer» Let’s take a to be the first term and d to be the common difference. And we know that, sum of first n terms Sn = \(\frac{n}{2}\)(2a + (n − 1)d) Given that sum of the first 7 terms of an A.P. is 63. S7 = 63 And sum of next 7 terms is 161. So, the sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms S14 = 63 + 161 = 224 Now, having S7 = \(\frac{7}{2}\)(2a + (7 − 1)d) ⟹ 63(2) = 7(2a + 6d) ⟹ 9 × 2 = 2a + 6d ⟹ 2a + 6d = 18 . . . . (1) And, S14 = \(\frac{14}{2}\)(2a + (14 − 1)d) ⟹ 224 = 7(2a + 13d) ⟹ 32 = 2a + 13d …. (2) Now, subtracting (1) from (2), we get ⟹ 13d – 6d = 32 – 18 ⟹ 7d = 14 ⟹ d = 2 Using d in (1), we have 2a + 6(2) = 18 2a = 18 – 12 a = 3 Thus, from nth term ⟹ a28 = a + (28 – 1)d = 3 + 27 (2) = 3 + 54 = 57 Therefore, the 28th term is 57. |
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511. |
Write the value of n for which nth of the A.P.s 3, 10, 17, ... and 63, 65, 67, ... are equal. |
Answer» Let nth term of A.P. 3,10,17,… be an And nth term of A.P. 63,65,67,…be an' So,a1 = 3 and a2 = 10 ∴ d = a2 - a1 = 10 - 3 = 7 And, a1' = 63 and a2' = 65 ∴ d' = a2' - a1' = 65 - 63 = 2 Now, an = a1+(n-1)d = 3+(n-1)7 = 3+7n-7 = 7n-4 And, an'=a1'+(n-1)d' = 63+(n-1)2 = 63+2n-2 = 2n+61 Now, It is given that an = an' ∴ 7n-4 = 2n+61 ∴ 7n-2n = 61+4 ∴ 5n = 65 ∴ n = 13 |
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512. |
For what value of n are the nth term of the following two A.P's the same. Also find this term 63, 65, 67, ... and 3, 10, 17, .... |
Answer» 1st AP = 63, 65, 67, ... Here, a = 63, d = 65 – 63 = 2 and 2nd AP = 3, 10, 17, ... Here, a = 3, d = 10 – 3 = 7 According to the question, 63 + (n – 1)2 = 3 + (n – 1)7 ⇒ 63 + 2n – 2 = 3 + 7n – 7 ⇒ 61 + 2n = 7n – 4 ⇒ 65 = 7n – 2n ⇒ 5n = 65 ⇒ n = 13 13th term of the given AP’s are same. Now, we will find the 13th term We have, an = a + (n – 1)d a13 = 63 + (13 – 1)2 a13 = 63 + 12 × 2 a13 = 63 + 24 a13 = 87 |
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513. |
In an A.P. the sum of first ten terms is -150 and the sum of its next 10 term is -550. Find the A.P. |
Answer» Let’s take a to be the first term and d to be the common difference. And we know that, sum of first n terms Sn = \(\frac{n}{2}\)(2a + (n − 1)d) Given that sum of the first 10 terms of an A.P. is -150. S10 = -150 And the sum of next 10 terms is -550. So, the sum of first 20 terms = Sum of first 10 terms + sum of next 10 terms S20 = -150 + -550 = -700 Now, having S10 = \(\frac{10}{2}\)(2a + (10 − 1)d) ⟹ -150 = 5(2a + 9d) ⟹ -30 = 2a + 9d ⟹ 2a + 9d = -30 . . . . (1) And, S20 = \(\frac{20}{2}\)(2a + (20 − 1)d) ⟹ -700 = 10(2a + 19d) ⟹ -70 = 2a + 19d …. (2) Now, subtracting (1) from (2), we get ⟹ 19d – 9d = -70 – (-30) ⟹ 10d = -40 ⟹ d = -4 Using d in (1), we have 2a + 9(-4) = -30 2a = -30 + 36 a = \(\frac{6}{2}\) = 3 Hence, we have a = 3 and d = -4 So, the A.P is 3, -1, -5, -9, -13,….. |
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514. |
Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term. |
Answer» Given, First term of the A.P is 1505 and S14 = 1505 We know that, the sum of first n terms is Sn = \(\frac{n}{2}\)(2a + (n − 1)d) So, S14 = \(\frac{14}{2}\)(2(10) + (14 − 1)d) = 1505 7(20 + 13d) = 1505 20 + 13d = 215 13d = 215 – 20 d = \(\frac{195}{13}\) d =15 Thus, the 25th term is given by a25 = 10 + (25 -1)15 = 10 + (24)15 = 10 + 360 = 370 Therefore, the 25th term of the A.P is 370. |
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515. |
For what value of n’. are the nth terms of two AP: 63, 65, 67, ……. and 3, 10, 17, ……….. equal? |
Answer» 63, 65, 67,………. a = 63. d = 65 – 63 = 2. an =? nth term of this is an = a + (n – 1)d = 63 + (n – 1) 2 = 63 + 2n – 2 an= 2n + 61 …………….. (i) 3, 10, 17, …………. a = 3, d = 10 – 3 = 7, an =? an = a + (n – 1)d = 3 + (n – 1)7 = 3 + 7n – 7 an = 7n — 4 ………….(ii) Here. nth terms of second AP are equal. ∴ equation (i) = equation (ii) 2n + 61 = 7n – 4 2n – 7n = -4 – 61 5n = 65 5n =65 ∴ n = \(\frac{65}{5}\) ∴ n = 13 ∴13th terms of the two given AP are equal. |
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516. |
Write the Properties of A.P. |
Answer» (A) For any real numbers a and b, the sequence whose nth term is an = an + b is always an A.P. with common difference ‘a’ (i.e. coefficient of term containing n) (B) If any nth term of sequence is a linear expression in n then the given sequence is an A.P. (C) If a constant term is added to or subtracted from each term of an A.P. then the resulting sequence is also an A.P. with the same common difference. (D) If each term of a given A.P. is multiplied or divided by a non-zero constant K, then the resulting sequence is also an A.P. with common difference Kd or......... respectively. Where d is the common difference of the given A.P. (E) In a finite A.P. the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of 1st and last term. (F) If three numbers a,b,c are in A.P., then 2b = a + c. |
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517. |
Check whether an = 2n2 + 1 is an A.p. or not. |
Answer» an = 2n2 + 1 Then an+1 = 2 (n + 1)2 + 1 So, an+1 - an = 2(n2 + 2n + 1) + 1 - 2n2 - 1 = 2n2 + 4n + 2 + 1 - 2n2 - 1 = 4n + 2, which is not constant So, The above sequence is not an A.P. |
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518. |
Kabir’s mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, “it seems like Kabir’s height grows in Arithmetic Progression”. Assuming this, she calculated how tall Kabir will be at the age of 15 years when he is in 10th! She was shocked to find it. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years. |
Answer» Height of Kabir when he was 1 year old = 70 cm Height of Kabir when he was 2 years old = 80 cm Height of Kabir when he was 3 years old = 90 cm The heights of Kabir form an A.P. Here, a = 70, d = 80 – 70 = 10 We have to find height of Kabir at the age of 15years i.e. t15 . Now, tn = a + (n – 1)d ∴ t15 = 70 + (15 – 1)10 = 70 + 14 × 10 = 70 + 140 ∴ t15 = 210 ∴ The height of Kabir at the age of 15 years will be 210 cm. |
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519. |
The 11th term of the AP: –5, (–5/2), 0, 5/2, …is(A) –20 (B) 20 (C) –30 (D) 30 |
Answer» (B) 20 Explanation: First term, a = – 5 Common difference, d = 5 – (-5/2) = 5/2 n = 11 We know that the nth term of an AP is an = a + (n – 1)d Where, a = first term an is nth term d is the common difference a11 = – 5 + (11 – 1)(5/2) a11 = – 5 + 25 = 20 |
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520. |
The first four terms of an AP, whose first term is –2 and the common difference is –2, are(A) – 2, 0, 2, 4(B) – 2, 4, – 8, 16(C) – 2, – 4, – 6, – 8(D) – 2, – 4, – 8, –16 |
Answer» (C) – 2, – 4, – 6, – 8 Explanation: First term, a = – 2 Second Term, d = – 2 a1 = a = – 2 We know that the nth term of an AP is an = a + (n – 1)d Where, a = first term an is nth term d is the common difference Hence, we have, a2 = a + d = – 2 + (- 2) = – 4 Similarly, a3 = – 6 a4 = – 8 So the A.P is – 2, – 4, – 6, – 8 |
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521. |
Find the sum of the first (i) 11 terms of the A.P : 2,6,10,14, ....(ii) 13 terms of the A.P : -6,0,6,12,....(iii) 51 terms of the A.P. whose second term is 2 and fourth term is 8. |
Answer» (i) a = 2, d = 6 – 2 = 4 S11=\(\frac{11}{2}\) [2(a) + 10d] = \(\frac{11}{2}\)[2(2) + 10(4)] = 11 [2 + 20] = 242 (ii) a = -6, d = 0 + 6 = 6 S13= \(\frac{13}{2}\)[2(a) + 12d] = 13 [-6 + 6(6)] = 13 [-6 + 36] = 13 (30) = 390 (iii) a2= 2 a + d = 2 (i) a4 = 8 a + 3d = 8 2 – d + 3d = 8 2 + 2d = 8 d = 3 a = -1 \(S_{21} = \frac{51}{2}[2(a) + 50d]\) = 51 [-1 + 25(3)] = 51 (74) = 3774 |
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522. |
Find the sum of the first:11 terms of the A.P. : 2, 6, 10, 14, . . . |
Answer» We know that the sum of terms for different arithmetic progressions is given by Sn = \(\frac{n}{2}\)[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms Given A.P 2, 6, 10, 14,… to 11 terms. Common difference (d) = a2 – a1 = 10 – 6 = 4 Number of terms (n) = 11 First term for the given A.P. (a) = 2 So, S11 = \(\frac{11}{2}\)[2(2) + (11 − 1)4] = \(\frac{11}{2}\)[2(2) + (10)4] = \(\frac{11}{2}\)[4 + 40] = 11 × 22 = 242 Hence, the sum of first 11 terms for the given A.P. is 242 |
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523. |
Write first four terms of the AP, when the first term ‘a’ and the common difference’d’ are given as follows: (i) a = 10, d = 10 (ii) a = -2, d = 0 (iii) a = 4, d = -3 (iv) a = -1, d = (v) a = -1.25, d = 0.25 |
Answer» (i) a = 10, d = 10 First four terms are, a, a + d, a + 2d, a + 3d 10, 10 + 10, 10 + 20, 10 + 30 10. 20, 30, 40. (ii) a = -2, d = 0 First four terms are, a, a + d, a + 2d, a + 3d -2. -2 + 0, -2 + 0, -2 + 0 -2. -2. -2, -2 (iii) a = 4, d = -3 First four terms are, a, a + d, a + 2d, a + 3d 4, 4 – 3, 4 – 2 × 3, 4 – 3 ×3 4, 1, -2, -5 iv) a = -1, d = \(\frac{1}{2}\) First four terms are, a, a + d, a + 2d, a + 3d - 1, - 1 + \(\frac{1}{2}, \) - 1 + 2 x \(\frac{1}{2}\), - 1 + 3 x \(\frac{1}{2}\) - 1, - \(\frac{1}{2}, \) 0, \(\frac{1}{2}, \) (v) a = -1.25, d = 0.25 First four terms are, a, a + d, a + 2d, a + 3d - 1.25, - 1.25 + 0.25, - 1.25 + 2 × 0.25, - 1.25 + 3 × 0.25 - 1.25, -1, - 0.75, - 0.50 |
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524. |
How many terms of the A.P. is 27, 24, 21. . . should be taken that their sum is zero? |
Answer» Given A.P. is 27, 24, 21. . . We know that, Sn = \(\frac{n}{2}\)[2a + (n − 1)d] Here we have, the first term (a) = 27 The sum of n terms (Sn) = 0 Common difference of the A.P. (d) = a2 – a1 = 24 – 27 = -3 On substituting the values in Sn, we get ⟹ 0 = \(\frac{n}{2}\)[2(27) + (n − 1)( − 3)] ⟹ 0 = (n)[54 + (n – 1)(-3)] ⟹ 0 = (n)[54 – 3n + 3] ⟹ 0 = n [57 – 3n] Further we have, n = 0 Or, 57 – 3n = 0 ⟹ 3n = 57 ⟹ n = 19 The number of terms cannot be zero, Hence, the numbers of terms (n) is 19. |
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525. |
How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693? |
Answer» Given A.P. is 63, 60, 57,… We know that, Sn = \(\frac{n}{2}\)[2a + (n − 1)d] Here we have, the first term (a) = 63 The sum of n terms (Sn) = 693 Common difference of the A.P. (d) = a2 – a1 = 60 – 63 = –3 On substituting the values in Sn we get ⟹ 693 = \(\frac{n}{2}\)[2(63) + (n − 1)(−3)] ⟹ 693 = \(\frac{n}{2}\)[163+(−3n + 3)] ⟹ 693 = \(\frac{n}{2}\)[129 − 3n] ⟹ 693(2) = 129n – 3n2 Now, we get the following quadratic equation. ⟹ 3n2 – 129n + 1386 = 0 ⟹ n2 – 43n + 462 Solving by factorisation method, we have ⟹ n2 – 22n – 21n + 462 = 0 ⟹ n(n – 22) -21(n – 22) = 0 ⟹ (n – 22) (n – 21) = 0 Either, n – 22 = 0 ⟹ n = 22 Or, n – 21 = 0 ⟹ n = 21 Now, the 22nd term will be a22 = a1 + 21d = 63 + 21( -3 ) = 63 – 63 = 0 So, the sum of 22 as well as 21 terms is 693. Therefore, the number of terms (n) is 21 or 22. |
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526. |
(5 + 13 + 21 + … + 181) = ? (a) 2476 (b) 2337 (c) 2219(d) 2139 |
Answer» Correct answer is (d) 2139 Here, a = 5, d = (13 - 5) = 8 and l = 181 Let the number of terms be n. Then, Tn = 181 \(\Rightarrow\) a + (n - 1) d = 181 \(\Rightarrow\) 5 + (n - 1)x8 = 181 \(\Rightarrow\) 8n = 184 \(\Rightarrow\) n = 23 ∴ Required sum = \(\frac{n}{2}(a+l)\) = \(\frac{23}{2}(5+181)=23\times93=2139\) Hence, the required sum is 2139. |
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527. |
Find the sum of first n natural numbers. |
Answer» For the given AP the first term a is 1, and common difference d is a difference of the second term and first term, which is 2 - 1 = 1 To find : the sum of given AP The formula for the sum of AP is given by, s = \(\frac{n}{2}\)(2a+(n-1)d) Substituting the values in the above formula, s = \(\frac{n}{2}\)(2+(n-1)(1)) s = \(\frac{n(n+1)}{2}\) |
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528. |
Find the sum of first n odd natural numbers. |
Answer» Given, An AP of first n odd natural numbers whose first term a is 1, and common difference d is 3 Given, Sequence is 1, 3, 5, 7……n To find : the sum of first n natural numbers Hence, The sum is given by the formula s = \(\frac{n}{2}\)(2a+(n-1)d) Substituting the values in the above equation we get, s = \(\frac{n}{2}\)(2+(n-1)2) s = n2 |
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529. |
If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero? |
Answer» Solution: Given, a3 = 4 and a9 = - 8 a3 = a + 2d = 4 a9 = a + 8d = - 8 Subtracting 3rd term from 9th term, we get; a + 8d – a – 2d = - 8 – 4 = - 12 Or, 6d = - 12 Or, d = - 2 Substituting the value of d in 3rd term, we get; a + 2(-2) = 4 Or, a – 4 = 4 Or, a = 8 Now; 0 = a + (n – 1)d Or, 0 = 8 + (n – 1)(- 2) Or, (n – 1)(- 2) = - 8 Or, n – 1 = 4 Or, n = 5 Thus, 5th term of this AP is zero. |
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530. |
Find the sum of all natural numbers between a and 100, which are divisible by 3. |
Answer» a = 3, l = 99, n = 33 S33 = \(\frac{33}{2} (a + l)\) = \(\frac{33}{2} (3 + 99)\) = \(\frac{33}{2}\) (3 + 99) = \(\frac{33}{2}\) (102) = 33 (51) = 1683 |
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531. |
If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms. |
Answer» Given sum up to n th term Sn= 4n-n^2 So n th term wil be tn = Sn - Sn-1= 4n -n^2-(4(n-1)-(n-1)^2 => tn= 4+2n -1=5-2n So sum of first 2 term S2= 4×2-2^2=4 t1 = 5-2×1=3 t2= 5- 2×2=1 t3 = 5-2×3=-1 t10 = 5 - 2×10=-15 |
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532. |
Find the sum of first 22 terms of an A.P. in which d = 22 and a22 = 149. |
Answer» a22= a + 21d 149 = a + 21(22) a = 149 – 462 a = -313 S22 = \(\frac{22}{2}\) [2a + (22 - 1)d] = 11 [-626 + 462] = 11 (-164) = -1804 |
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533. |
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. |
Answer» d = 7, a22 = 149, S22 = ? a = a + (n – 1)d a + (22 – 1)7 = 149 a + 21 × 7 = 149 a + 147 = 149 =149 – 147 ∴ a = 2 ∴ Sn = \(\frac{n}{2}\)[a + an] S22 = \(\frac{22}{2}\)[2 + 149] = 11[151] ∴ S22 = 1661 |
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534. |
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. |
Answer» Solution: Given d = 7, |
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535. |
Define an arithmetic progression. |
Answer» An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. This fixed number is called the common difference of the AP. |
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536. |
Find the 20th term from the Last term of the AP: 3, 8, 13, …………., 253. |
Answer» 3, 8, 13, ………….., 253 a = 3. d = 8 – 3 = 5, an = 253 20th term from the last term of the AP starting from 253 =? 253, 258, 263, ………… a20 =? a = 253, d = 258 – 253 = 5, n = 20 an = a + (n – 1)d a20= 253 + (20 – 1) 5 = 253 + 19 × 5 = 253 + 95 ∴ a20 = 348 ∴ 20th term from the last term of the AP is 348. |
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537. |
Find the sum of first 22 terms of an A.P. in which d = 22 and a22 = 149. |
Answer» Let the first term be taken as a. Given, a22 = 149 and the common difference d = 22 Also, we know that an = a + ( n – 1) d So, the 22nd term is given by a22 = a + (22 – 1)d 149 = a + (21) (22) a = 149 – 462 a = – 313 Now, for the sum of term Sn = \(\frac{n}{2}\)[2a + (n − 1)d] Here, n = 22 S22 = \(\frac{22}{2}\)[2(−313) + (22 − 1)(22)] = (11)[ – 626 + 462] = (11)[–164] = – 1804 Hence, the sum of first 22 terms for the given A.P. is S22 = -1804 |
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538. |
The nth term of an A.P., the sum of whose n terms is Sn, is A. Sn + Sn–1 B. Sn – Sn–1 C. Sn + Sn + 1 D. Sn – Sn + 1 |
Answer» The sum of n terms of an A.P is given by Sn Sn = \(\frac{n}{2}\)(2a + (n–1) d) If the sum of n terms is given that is Sn is given then the nth term is given by the formula Tn = Sn– Sn–1 Where Sn–1 is sum of the (n–1)th term of the A.P. |
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539. |
Write the common difference of an A.P. whose nth term is an = 3n + 7. |
Answer» If an = 3n + 7 Then a1 = 3 (1) + 7 = 10 a2 = 3(2) + 7 = 13 a3 = 3(3) + 7 = 16 d= common difference = an – an–1 = a3 – a2 = 16 – 13 = 3 |
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540. |
30th term of the AP: 10, 7, 4, ……….. is (A) 97 (B) 77 (C) -77 (D) -87 |
Answer» Answer is (C) a = 10. d = 7 – 10 = -3, n = 30, a30 =? an = a + (n – 1)d a30 = 10 + (30 – 1)(-3) = 10 + 29(-3) = 10 – 87 ∴ a30 = -77 |
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541. |
Write the nth term of an A.P. the sum of whose n terms is Sn. |
Answer» First term = a Sum up to first term = a Last term (nth term) = an Sum up to n terms = Sn Second last term = an–1 Sum up to (n–1)th term = sn–1 Therefore, an = Sn–Sn–1 |
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542. |
Which item of the sequence 114, 109, 104, .... is the first negative term? |
Answer» Here a = 114, d is common difference d = a3 – a2 = a2 – a1 = –5 For finding first negative term Tn < 0 a + (n – 1) d < 0 114 + (n – 1) (– 5) < 0 114 – 5n + 5 < 0 119 – 5n < 0 – 5n < – 119 5n > 119 n > 119/5 n > 23.8 Therefore first negative term is 24th term |
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543. |
If the nth term of the A.P. 9, 7, 5, …. is same as the nth term of the A.P. 15, 12, 9, … find n. |
Answer» Given, A.P1 = 9, 7, 5, …. and A.P2 = 15, 12, 9, … And, we know that, nth term an = a + (n – 1)d For A.P1, a = 9, d = Second term – first term = 9 – 7 = -2 And, its nth term an = 9 + (n – 1)(-2) = 9 – 2n + 2 an = 11 – 2n …..(i) Similarly, for A.P2 a = 15, d = Second term – first term = 12 – 15 = -3 And, its nth term an = 15 + (n – 1)(-3) = 15 – 3n + 3 an = 18 – 3n …..(ii) According to the question, its given that nth term of the A.P1 = nth term of the A.P2 ⇒ 11 – 2n = 18 – 3n n = 7 Therefore, the 7th term of the both the A.Ps are equal. |
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544. |
Find the 12th term from the end of the arithmetic progressions:1, 4, 7, 10, … ,88 |
Answer» Given A.P = 1, 4, 7, 10, … ,88 Here, a = 1 and d = (4 – 1) = 3 Now, find the number of terms when the last term is known i.e, 88 an = 1 + (n – 1)3 = 88 1 + 3n – 3 = 88 3n = 90 n = 30 Hence, the A.P has 30 terms. So, the 12th term from the end is same as (30 – 12 + 1)th of the A.P which is the 19th term. ⇒ a89 = 1 + (19 – 1)3 = 1 + 18(3) = 1 + 54 = 55 Therefore, the 12th term from the end of the A.P is 55. |
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545. |
Find the 12th term from the end of the arithmetic progressions:3,8,13, … ,253 |
Answer» Given A.P = 3,8,13, … ,253 Here, a = 3 and d = (8 – 3) = 5 Now, find the number of terms when the last term is known i.e, 253 an = 3 + (n – 1)5 = 253 3 + 5n – 5 = 253 5n = 253 + 2 = 255 n = 255/5 n = 51 Hence, the A.P has 51 terms. So, the 12th term from the end is same as (51 – 12 + 1)th of the A.P which is the 40th term. ⇒ a40 = 3 + (40 – 1)5 = 3 + 39(5) = 3 + 195 = 198 Therefore, the 12th term from the end of the A.P is 198. |
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546. |
Write the value of a30 – a10 for the A.P. 4, 9, 14, 19, ………. |
Answer» Here a1 = 4 a2 = 9 a3 = 14 d = common difference = a3 – a2 = a2 – a1 = 14 – 9 = 9 – 4 = 5 a30 = a + (n–1) d a30 = 4 + (30 –1) 5 a30= 4 + 29 X 5 a30= 149 Similarly, a10 = 4 + 9x5 a10= 49 a30 – a10 = 149 – 49 = 100 |
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547. |
Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P. |
Answer» Terms are in A.P. if common difference (d) is same between two continuous numbers. x + 10 – 2x = 3x + 2 – (x + 10) 10 – x = 2x – 8 18 = 3x x= 6 So the terms in A.P. are: 2(6), 6 + 10, 3(6) + 2 = 12, 16, 20 |
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548. |
Find the 12th term from the end of the arithmetic progressions:3, 5, 7, 9, …. 201 |
Answer» Given A.P = 3, 5, 7, 9, …. 201 Here, a = 3 and d = (5 – 3) = 2 Now, find the number of terms when the last term is known i.e, 201 an = 3 + (n – 1)2 = 201 3 + 2n – 2 = 201 2n = 200 n = 100 Hence, the A.P has 100 terms. So, the 12th term from the end is same as (100 – 12 + 1)th of the A.P which is the 89th term. ⇒ a89 = 3 + (89 – 1)2 = 3 + 88(2) = 3 + 176 = 179 Therefore, the 12th term from the end of the A.P is 179. |
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549. |
The 8th term of an A.P. is 37, and its 12th term is 57. Find the A.P. |
Answer» We have a8 = a + (8 – 1)d = a + 7d = 37 …(1) and a12 = a + (12 – 1)d = a + 11d = 57 …(2) Solving the pair of linear equations (1) and (2), we get a + 7d – a – 11d = 37 – 57 ⇒ – 4d = –20 ⇒ d = 5 Putting the value of d in eq (1), we get a + 7(5) = 37 ⇒ a + 35 = 37 ⇒ a = 2 Hence, the required AP is 2, 7, 12, 17,… |
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550. |
Write 5th term from the end of the A.P. 3, 5, 7, 9, ...., 201. |
Answer» Here a = 201, a2 = 5, a3 = 7 d = a3 – a2 = a2 – a1 = 7 – 5 = 5 – 3 = 2 tn = a + (n –1)d tn = 3 + (n–1)2 = 201 (n –1)2 = 201 – 3 = 198 n – 1 =\(\frac{198}{2}\) = 99 n = 99 + 1 =100 5th term from end = 96th term T95 = 3 + (96 –1)2 T95 = 3 + 95 x 2 T95 = 3 + 190 = 193 |
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