Q1: Find the sum of all multiples of 7 lying between 500 and 900.
Ans: 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
Q2: The sums of first n terms of three arithmetic progressions are S_{1 }S_{2} and S_{3} respectively. The first term of each A.P. is 1 and their common differences are 1, 2 and 3 respectively. Prove that S_{1} + S_{3} = 2S_{2}.
Ans:
Q3: 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.
Ans: 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
Q4: If the ratio of the sum of first n terms of two A.Ps is (7n + 1): (4n + 27), find the ratio of their mth terms.
Ans: Let A be first term and D be the common difference of 1st A.P.
Let a be the first term and d be the common difference of 2nd A.P.
∴ Required ratio = (14m – 6) : (8m + 23)
Q5: If S_{n} denotes the sum of first n terms of an A.P., prove that S_{30} = 3[S_{20} – S_{10}].
Ans: Let a be the first term and d be the common difference of the A.P.
= 10a + 45d …(iii)
To prove, 3(S_{20} – S_{10}) = S_{30}
= 3(20a + 190d – 10a – 45d) …[From (i), (ii) & (iii)
= 3(10a + 145d)
= 30a + 435d = S30 …Hence Proved
Q6: If the sum of the first n terms of an A.P. is 1/2 (3n^{2} + 7n), then find its nth term. Hence write its 20th.
Ans:
Now we know,
S_{1} = 27 and a_{2} = S_{2} – S_{1}
∴ a_{1 }= 5 = 13 – 5 = 8
Now we have,
a_{1} = 5, a_{2 }= 8, d = a_{2} – a_{1} = 8 – 5 = 3
an = a + (n – 1)d = 5 + (n – 1)3
= 5 + 3n – 3 = 3n + 2 (nth term)
∴ 20th term, a_{20} = (3 × 20) + 2 = 62
Q7: If Sn, denotes the sum of first n terms of an A.P., prove that S_{12} = 3(S_{8} – S_{4}).
Ans: Let a be the first term and d be the common difference of A.P.
12a + 660 = 3(8a + 28d – 4a – 6d) … [From (i), (ii) & (iii)
12a + 660 = 3(4a + 22d)
12a + 660 = 12a + 66d …Hence proved
Q8: The sum of the first seven terms of an AP is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the AP.
Ans: S_{7} = 182 …[Given]
Q9: The sum of first n terms of an AP is 3n^{2} + 4n. Find the 25^{th} term of this AP.
Ans: 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
Q10: The n^{th} term of an A.P. is given by (-4n + 15). Find the sum of first 20 terms of this A.P.
Ans: We have, a_{n} = -4n + 15
Put n = 1, a_{1} = -4(1) + 15 = 11
Put n=2, a_{2} = -4(2) + 15 = 7
∴ d = a_{2} – a_{1} = 7 – 11 = -4
As S_{n} = n/2 (2a + (n – 1)d]
∴ S_{20} = 20/2[2(11) + (20 – 1)(-4))… [n = 20 (Given)]
= 10 (22 – 76)
= 10 (-54) = -540
Q11: 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.
Ans: First term, a = 5, Last term, a_{n} = 45
Let the number of terms = n
S_{n} = 400
Q12: If the sum of first 7 terms of an A.P is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P.
Ans:
Let 1^{st} term = a, Common difference = d
Given: S_{7} = 49, S_{17} = 289
Q13: The 13th term of an AP is four times its 3rd term. If its fifth term is 16, then find the sum of its first ten terms.
Ans:
a_{13} = 4a_{3} … [Given
⇒ a + 12d = 4(a + 2d) …[∵ a_{n} = a + (n – 1)d
⇒ a + 12d – 4a – 8d = 0
Q14: If 3k-2, 4k-6 and k+2 are three consecutive terms of AP, then find the value of k.
Ans:
To be term of an AP the difference between two
consecutive terms must be the same.
If 3k-2, 4k-6 and k+2 are terms of an AP, then
4k -6 -(3k-2) = k + 2 - (4k-6)
⇒ 4k - 6 -3k + 2 = k + 2 -4k + 6
⇒ k - 4 = 8 - 3k
⇒ 4k = 12
⇒ k = 3
Hence, the value of k is 3.
Q15: Find the number of terms of the AP 18, 151/2, 13,…, -491/2 and find the sum of all its terms.
Ans: Here 1^{st} term, a = 18
Q16: Find the value of the middle term of the following A.P.: -6, -2, 2, …, 58.
Ans: Here a = -6, d = -2 -(-6) = 4, a_{n} = 58
As we know, a + (n – 1) d = 58
∴ -6 + (n – 1) 4 = 58
⇒ (n – 1) 4 = 58 + 6 = 64
⇒ (n – 1) = 64/4 = 16
⇒ n = 16 + 1 = 17 (odd)
Q17: If the seventh term of an AP is 1/9 and its ninth term is 1/7, find its 63rd term.
Ans:
Q18: 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.
Ans:
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
Q19: The 9th term of an A.P. is equal to 6 times its second term. If its 5^{th} term is 22, find the A.P.
Ans:
9^{th} term = 6 (2^{nd} term)
∴ a +8d = 6 (a + d) …[As a_{n} = a + (n – 1)d
a + 8d = 6a + 6d
8d – 6d = 6a – a
2d = 5a
⇒ d = 5a/2 …(i)
Now, a_{5} = 22
a + 4d = 22
Q20: 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.
Ans:
Given: a_{19} = 3(a_{6})
⇒ a + 18d = 3(a + 5d)
a + 18d = 3a + 15d
18d – 15d = 3a – a