In an AP, if \(d=-4\), \(n=7\), \(a_n=4\), then \(a\) is
6
7
20
28
In an AP, if \(a=3.5\), \(d=0\), \(n=101\), then \(a_n\) will be
0
3.5
103.5
104.5
The list of numbers \(-10,-6,-2,2,\ldots\) is
an AP with \(d=-16\)
an AP with \(d=4\)
an AP with \(d=-4\)
not an AP
The 11th term of the AP \(-5,\; -\dfrac{5}{2},\; 0,\; \dfrac{5}{2},\ldots\) is
−20
20
−30
30
The first four terms of an AP with \(a=-2\) and \(d=-2\) are
−2, 0, 2, 4
−2, 4, −8, 16
−2, −4, −6, −8
−2, −4, −8, −16
The 21st term of the AP whose first two terms are \(-3\) and \(4\) is
17
137
143
−143
If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
30
33
37
38
Which term of the AP \(21,42,63,84,\ldots\) is \(210\)?
9th
10th
11th
12th
If the common difference of an AP is 5, then what is \(a_{18}-a_{13}\)?
5
20
25
30
What is the common difference of an AP in which \(a_{18}-a_{14}=32\)?
8
−8
−4
4
Two APs have the same common difference. The first term of one is \(-1\) and of the other is \(8\). Then the difference between their 4th terms is
−1
−8
7
−9
If \(7\) times the 7th term of an AP is equal to \(11\) times its 11th term, then its 18th term will be
7
11
18
0
The 4th term from the end of the AP: \(-11,-8,-5,\ldots,49\) is
37
40
43
58
The famous mathematician associated with finding the sum of the first 100 natural numbers is
Pythagoras
Newton
Gauss
Euclid
If the first term of an AP is \(-5\) and the common difference is \(2\), then the sum of the first 6 terms is
0
5
6
15
The sum of first 16 terms of the AP: \(10,6,2,\ldots\) is
−320
320
−352
−400
In an AP if \(a=1\), \(a_n=20\) and \(S_n=399\), then \(n\) is
19
21
38
42
The sum of first five multiples of 3 is
45
55
65
75
Which of the following form an AP? Justify your answer.
(i) \(-1, -1, -1, -1, \ldots\)
(ii) \(0, 2, 0, 2, \ldots\)
(iii) \(1, 1, 2, 2, 3, 3, \ldots\)
(iv) \(11, 22, 33, \ldots\)
(v) \(\dfrac{1}{2},\; \dfrac{1}{3},\; \dfrac{1}{4},\; \ldots\)
(vi) \(2, 2^2, 2^3, 2^4, \ldots\)
(vii) \(\sqrt{3},\; \sqrt{12},\; \sqrt{27},\; \sqrt{48},\; \ldots\)
(i) Yes (\(d=0\)); (ii) No; (iii) No; (iv) Yes (\(d=11\)); (v) No; (vi) No; (vii) Yes (\(d=\sqrt{3}\)).
Is it true that \(-1,\; -\dfrac{3}{2},\; -2,\; -\dfrac{5}{2},\; \ldots\) forms an AP because \(a_2-a_1=a_3-a_2\)? Justify.
True.
For the AP \(-3,-7,-11,\ldots\), can we find \(a_{30}-a_{20}\) directly without actually finding \(a_{30}\) and \(a_{20}\)? Explain.
Yes. \(a_{30}-a_{20}=(30-20)\,d=10(-4)=-40\).
Two APs have the same common difference. The first term of one AP is \(2\) and that of the other is \(7\). Show that the difference between their 10th terms equals the difference between their 21st terms (indeed, between any two corresponding terms). Why?
The difference is constant and equals \(2-7=-5\) for every corresponding term.
Is \(0\) a term of the AP \(31, 28, 25, \ldots\)? Justify your answer.
No.
The taxi fare after each km, when the fare is Rs 15 for the first km and Rs 8 for each additional km, does not form an AP as the total fare (in Rs) after each km is \(15, 8, 8, 8, \ldots\). Is this statement true? Give reasons.
False. The total fare after each km is \(15, 23, 31, 39, \ldots\), which is an AP with \(a=15\), \(d=8\).
In which of the following situations do the lists of numbers form an AP? Give reasons.
(i) The fee charged every month by a school for a whole session when the monthly fee is Rs 400.
(ii) The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is Rs 250, and it increases by Rs 50 for the next higher class.
(iii) The amount of money in Varun’s account at the end of every year when Rs 1000 is deposited at simple interest of 10% p.a.
(iv) The number of bacteria after each second when they double every second.
(i) Yes (\(d=0\)); (ii) Yes (\(d=50\)); (iii) Yes (\(d=100\)); (iv) No (geometric, not arithmetic).
Decide whether each expression can be the \(n\)th term of an AP. Justify.
(i) \(2n-3\)
(ii) \(3n^2+5\)
(iii) \(1+n+n^2\)
(i) Yes; (ii) No; (iii) No.
Match the APs in Column A with their common differences in Column B.
Column A:
(A1) 2, −2, −6, −10, …
(A2) \(a = -18,\; n = 10,\; a_n = 0\)
(A3) \(a = 0,\; a_{10} = 6\)
(A4) \(a_2 = 13,\; a_4 = 3\)
Column B:
(B1) \(\dfrac{2}{3}\), (B2) −5, (B3) 4, (B4) −4, (B5) 2, (B6) \(\dfrac{1}{2}\), (B7) 5
Matches: (A1)→(B4), (A2)→(B5), (A3)→(B1), (A4)→(B2).
Verify each is an AP and write its next three terms.
(i) \(0,\; \dfrac{1}{4},\; \dfrac{1}{2},\; \dfrac{3}{4},\ldots\)
(ii) \(5,\; \dfrac{14}{3},\; \dfrac{13}{3},\; 4,\ldots\)
(iii) \(\sqrt{3},\; 2\sqrt{3},\; 3\sqrt{3},\ldots\)
(iv) \(a+b,\; (a+1)+b,\; (a+1)+(b+1),\ldots\)
(v) \(a,\; 2a+1,\; 3a+2,\; 4a+3,\ldots\)
(i) Yes, \(d=\dfrac{1}{4}\). Next: \(1,\; \dfrac{5}{4},\; \dfrac{3}{2}\).
(ii) Yes, \(d=-\dfrac{1}{3}\). Next: \(\dfrac{11}{3},\; \dfrac{10}{3},\; 3\).
(iii) Yes, \(d=\sqrt{3}\). Next: \(4\sqrt{3},\;5\sqrt{3},\;6\sqrt{3}\).
(iv) Yes, \(d=1\). Next: \(a+b+3,\; a+b+4,\; a+b+5\).
(v) Yes, \(d=a+1\). Next: \(5a+4,\; 6a+5,\; 7a+6\).
Write the first three terms of the AP given \(a\) and \(d\).
(i) \(a=\dfrac{1}{2},\; d=-\dfrac{1}{6}\)
(ii) \(a=-5,\; d=-3\)
(iii) \(a=\sqrt{2},\; d=\dfrac{1}{\sqrt{2}}\)
(i) \(\dfrac{1}{2},\; \dfrac{1}{3},\; \dfrac{1}{6}\)
(ii) \(-5,\; -8,\; -11\)
(iii) \(\sqrt{2},\; \sqrt{2}+\dfrac{1}{\sqrt{2}},\; \sqrt{2}+\dfrac{2}{\sqrt{2}}=2\sqrt{2}\)
Find \(a, b, c\) such that \(a,\;7,\;b,\;23,\;c\) are in AP.
\(a=-1,\; b=15,\; c=31\)
Determine the AP whose 5th term is 19 and \(a_{13}-a_{8}=20\).
\(a=3,\; d=4\) (AP: 3, 7, 11, …)
The 26th, 11th and last terms are \(0,\;3\) and \(-\dfrac{1}{5}\), respectively. Find the common difference and number of terms.
\(d=-\dfrac{1}{5},\; n=27\)
\(a_5+a_7=52\) and \(a_{10}=46\). Find the AP.
\(a=1,\; d=5\) (AP: 1, 6, 11, …)
Find \(a_{20}\) if \(a_1=12\) and \(a_7\) is 24 less than \(a_{11}\).
\(a_{20}=126\)
Is 55 a term of AP \(7,10,13,\ldots\)? If yes, which term?
Yes, the 17th term.
Determine \(k\) so that \(k^2+4k+8\), \(2k^2+3k+6\), \(3k^2+4k+4\) are three consecutive terms of an AP.
\(k=0\)
Split 207 into three parts in AP such that the product of the two smaller parts is 4623.
67, 69, 71
The angles of a triangle are in AP; the greatest is twice the least. Find all angles.
40°, 60°, 80°
If the \(n\)th terms of APs \(9,7,5,\ldots\) and \(24,21,18,\ldots\) are equal, find \(n\) and that term.
\(n=16\), term = \(-21\)
If \(a_3+a_8=7\) and \(a_7+a_{14}=-3\), find \(a_{10}\).
\(a_{10}=-1\)
Find the 12th term from the end of AP: \(-2,-4,-6,\ldots,-100\).
\(-78\)
Which term of AP \(53,48,43,\ldots\) is the first negative term?
12th
How many numbers between 10 and 300 leave remainder 3 when divided by 4?
73
Find the sum of the two middle-most terms of the AP: \(-\dfrac{4}{3}, -1, -\dfrac{2}{3},\ldots, 4\dfrac{1}{3}\).
3
First term is \(-5\), last term is \(45\), and sum of the AP is \(120\). Find the number of terms and the common difference.
\(n=6,\; d=10\)
Find the sum:
(i) \(1+(-2)+(-5)+\cdots+(-236)\)
(ii) \(\big(4-\dfrac{1}{n}\big)+\big(4-\dfrac{2}{n}\big)+\cdots+\big(4-\dfrac{n}{n}\big)\)
(iii) \(\dfrac{a-b}{a+b}+\dfrac{3a-2b}{a+b}+\dfrac{5a-3b}{a+b}+\cdots\) to 11 terms
(i) \(-9400\)
(ii) \(\dfrac{7n-1}{2}\)
(iii) \(\dfrac{121a-66b}{a+b}\)
Which term of the AP \(-2,-7,-12,\ldots\) is \(-77\)? Find the sum up to that term.
16th term; sum = −632
If \(a_n=3-4n\), show \(a_1,a_2,a_3,\ldots\) form an AP and find \(S_{20}\).
AP with \(d=-4\); \(S_{20}=-780\)
In an AP, if \(S_n=n(4n+1)\), find the AP.
AP: 5, 13, 21, … (\(a=5, d=8\))
In an AP, if \(S_n=3n^2+5n\) and \(a_k=164\), find \(k\).
\(k=27\)
If \(S_n\) is the sum of first \(n\) terms of an AP, prove \(S_{12}=3(S_8-S_4)\).
Proved.
Find \(S_{17}\) if \(a_4=-15\) and \(a_9=-30\).
\(S_{17}=-510\)
If \(S_6=36\) and \(S_{16}=256\), find \(S_{10}\).
\(S_{10}=100\)
Find the sum of all 11 terms of an AP whose middle-most term is 30.
330
Find the sum of the last 10 terms of the AP: \(8,10,12,\ldots,126\).
1170
Find the sum of first seven numbers which are multiples of 2 as well as of 9.
504
How many terms of the AP: \(-15,-13,-11,\ldots\) are needed to make the sum \(-55\)? Explain the double answer.
5 terms or 11 terms
The sum of first \(n\) terms of an AP with \(a=8\), \(d=20\) equals the sum of first \(2n\) terms of another AP with \(a=-30\), \(d=8\). Find \(n\).
\(n=11\)
Kanika deposits Rs 1 on day 1, Rs 2 on day 2, … through January 2008 (31 days). She also spent Rs 204 and still had Rs 100 left. What was her pocket money for the month?
Rs 800
Yasmeen saves Rs 32, Rs 36, Rs 40 in months 1,2,3 and continues so (AP with \(d=4\)). In how many months will she save Rs 2000?
25 months
The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms is 235, find the sum of its first twenty terms.
970
Find the
(i) sum of integers from 1 to 500 that are multiples of 2 as well as 5,
(ii) sum of integers from 1 to 500 that are multiples of 2 as well as 5 (inclusive),
(iii) sum of integers from 1 to 500 that are multiples of 2 or 5.
(i) 12750, (ii) 12750, (iii) 75250
The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.
3
An AP consists of 37 terms. The sum of the three middle-most terms is 225 and the sum of the last three is 429. Find the AP.
First term \(a=3\), common difference \(d=4\) (AP: 3, 7, 11, …)
Find the sum of integers between 100 and 200 that are
(i) divisible by 9, (ii) not divisible by 9.
(i) 1683 (ii) 13167
The ratio of the 11th term to the 18th term of an AP is \(2:3\). Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.
\(a_5:a_{21}=1:3\), \(S_5:S_{21}=5:49\)
Show that the sum of an AP whose first term is \(a\), second term is \(b\) and last term is \(c\), is
\[ S = \dfrac{(a+c)(b+c-2a)}{2(b-a)}. \]
\(S=\dfrac{(a+c)(b+c-2a)}{2(b-a)}\)
Solve the equation \(-4+(-1)+2+\cdots+x=437\).
\(x=50\)
Jaspal Singh repays a loan of Rs 118000 by monthly instalments starting with Rs 1000 and increasing by Rs 100 every month. What will he pay in the 30th instalment? What amount is still unpaid after the 30th instalment?
30th instalment: Rs 3900; Unpaid after 30th: Rs 44,500
Along a straight passage, 27 flags are to be fixed at intervals of 2 m. The flags are stored at the position of the middle-most flag. Ruchi places the flags, carrying only one at a time and returning to the store each time. How much total distance does she cover to complete the job and return? What is the maximum distance she travels while carrying a flag?
Total distance: 364 m; Maximum carrying distance: 26 m