Mathematics Class X

Question 1 of 30

1. Question
1 point(s)
In an AP, if a=28, d=-3, n=6 then ${a_n}$is?
- 4
- 5
- 8
- 13
Correct

Incorrect

Hint

$\begin{array}{l}{a_n} = a + (n - 1)d\\ = 28 + (6 - 1)( - 3)\\ = 28 + ( - 15)\\ = 13\end{array}$
Question 2 of 30

2. Question
1 point(s)
If a=20, d=15 then first four terms will be?
- 20,35,50,65
- 20,30,40,50
- 15,30,45,60
- 10,20,30,40
Correct

Incorrect

Hint

$\begin{array}{l}{a_1} = 20\\{a_2} = 20 + 15 = 35\\{a_3} = 35 + 15 = 50\\{a_4} = 50 + 15 = 65\end{array}$
Question 3 of 30

3. Question
1 point(s)
The first term and common difference for the AP 3, 1,-1,-3 are?
- 1 and 3
- -1 and 3
- 3 and -2
- 2 and 3
Correct

Incorrect

Hint

$\begin{array}{l}{a_3} = 3\\d = 1 - 3 = - 2\end{array}$
Question 4 of 30

4. Question
1 point(s)
${29^{th}}$Term of the AP 10,7,4… is?
- -74
- -100
- -96
- -95
Correct

Incorrect

Hint

$\begin{array}{l}{a_{29}} = a + 28d\\ = 10 + 28 \times ( - 3)\\ = 10 - 84\\ = - 74\end{array}$
Question 5 of 30

5. Question
1 point(s)
10th term of the AP -3,$\dfrac{{ – 1}}{2}$, 2… is?
- 39
- $\dfrac{{39}}{2}$
- 15
- 19
Correct

Incorrect

Hint

$\begin{array}{l}d = \dfrac{{ - 1}}{2} - ( - 3) = \dfrac{{ - 1}}{2} + 3\\ = \dfrac{5}{2}\end{array}$
$\begin{array}{l}{a_{10}} = a + 9d\\ = - 3 + 9 \times \dfrac{5}{2}\\ = - 3 + \dfrac{{45}}{2}\\ = \dfrac{{ - 6 + 45}}{2} = \dfrac{{39}}{2}\end{array}$
Question 6 of 30

6. Question
1 point(s)
The missing term in AP _, 13, _3 are?
- 11 and 9
- 17 and 9
- 18 and 8
- 18 and 9
Correct

Incorrect

Hint

$\begin{array}{l}{a_2} = 13,{a_4} = 3\\{a_2} = a + (2 - 1)d\\13 = a + d......(1)\\{a_4} = a + 3d\\3 = a + 3d......(2)\end{array}$
Substituting equation 1 from 2 we get
$\begin{array}{l} - 10 = 2d \Rightarrow d = - 5\\a + d = 13\\ \Rightarrow a + ( - 5) = 13\\ = a = 18\\{a_3} = a + 2d\\ = 18 + 2( - 5) = 8\end{array}$
Question 7 of 30

7. Question
1 point(s)
If an AP has a=1, ${t_n} = 20$and${S_n} = 399$, then value of n is?
- 20
- 32
- 38
- 40
Correct

Incorrect

Hint

$\begin{array}{l}{S_n} = \dfrac{1}{2}\left[ {(a + t)} \right] \times n\\ \Rightarrow 399 = (1 + 20) \times \dfrac{n}{2}\\ \Rightarrow 399 \times 2 = 21n\\ \Rightarrow n = \dfrac{{399 \times 2}}{{21}} = 38\end{array}$
Question 8 of 30

8. Question
1 point(s)
Which term of the AP 92, 88, 84, 80, 76… is 0?
- 23
- 32
- 22
- 24
Correct

Incorrect

Hint

$\begin{array}{l}{a_n} = 0\\ \Rightarrow 0 = 92 + (n - 1) \times ( - 4)\\ \Rightarrow 0 = 92 - 4n + 4\\ \Rightarrow n = 24\end{array}$
Question 9 of 30

9. Question
1 point(s)
In an AP if d=-4, n=6, ${a_n} = 4$then a=?
- 6
- 24
- 32
- 36
Correct

Incorrect

Hint

$\begin{array}{l}{a_n} = a + (n - 1)d\\4 = a + 5 \times ( - 4)\\4 + 20 = a\\a = 24\end{array}$
Question 10 of 30

10. Question
1 point(s)
The 10th term of an AP $ – 5,\dfrac{{ – 5}}{2},0,\dfrac{5}{2},….$is?
- 20
- 25
- 35
- $\dfrac{{35}}{2}$
Correct

Incorrect

Hint

$\begin{array}{l}d = \dfrac{{ - 5}}{2} + 5 = \dfrac{5}{2}\\{a_{10}} = a + 9d\\ = - 5 + 9 \times \dfrac{5}{2}\\ = - 5 + \dfrac{{45}}{2}\\ = \dfrac{{ - 10 + 45}}{2} = \dfrac{{35}}{2}\end{array}$
Question 11 of 30

11. Question
1 point(s)
The 18th term of an AP whose first two terms are -3 and 4 is?
- 116
- 120
- 121
- 143
Correct

Incorrect

Hint

$\begin{array}{l}{a_1} = - 3,{a_2} = 4\\d = 7\\{a_{18}} = a + 17d\\ = - 3 + 17 \times 7\\ = - 3 + 119\\ = 116\end{array}$
Question 12 of 30

12. Question
1 point(s)
Which term of an AP 21, 42, 63, 84, are 210?
- 9th
- 11th
- 15th
- 10th
Correct

Incorrect

Hint

$\begin{array}{l}{a_n} = a + (n - 1)d\\210 = 21 + (n - 1)21\\210 = 21 + 21n - 21\\n = 10\end{array}$
Question 13 of 30

13. Question
1 point(s)
If the common difference is 6 then the value of${a_{17}} – {a_{12}}$?
- 30
- 25
- 20
- 15
Correct

Incorrect

Hint

$\begin{array}{l}{a_{17}} - {a_{12}} = (a + 16d) - (a + 11d)\\ = a + 16d - a - 11d\\ = 5d = 5 \times 6 = 30\end{array}$
Question 14 of 30

14. Question
1 point(s)
If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be?
- 0
- 1
- 2
- 3
Correct

Incorrect

Hint

$\begin{array}{l}7{a_7} = 11{a_{11}}\\7(a + 6d) = 11(a + 10d)\\ \Rightarrow 7a + 42d = 11a + 110d\\ \Rightarrow 4a + 68d = 0\\ \Rightarrow 2\left( {2a + 34d} \right) = 0\\ \Rightarrow a + 17d = 0\end{array}$
Question 15 of 30

15. Question
1 point(s)
The sum of first 15 terms of the AP 10, 6, -2 … is?
- -270
- -250
- 320
- 270
Correct

Incorrect

Hint

$\begin{array}{l}{s_{15}} = \dfrac{{15}}{2}\left[ {2a + 14d} \right]\\ = \dfrac{{15}}{2}\left[ {2 \times 10 + 14 \times ( - 4)} \right]\\ = \dfrac{{15}}{2}\left[ {20 - 56} \right]\\ = \dfrac{{15}}{2} \times ( - 36) - 15 \times ( - 18)\\ = - 270\end{array}$
Question 16 of 30

16. Question
1 point(s)
The sum of first 5 multiples of 3 is?
- 45
- 55
- 75
- 95
Correct

Incorrect

Hint

$\begin{array}{l}{S_5} = \dfrac{5}{2}\left[ {2a + 4d} \right]\\ = \dfrac{5}{2}\left[ {2 \times 3 + 4 \times 3} \right]\\ = \dfrac{5}{2}\left[ {6 + 12} \right] = 5 \times = 45\end{array}$
Question 17 of 30

17. Question
1 point(s)
The value of a, b, c such that the following numbers are in AP a, 7, b, 23, c
- -1,2,30
- -1,15,31
- 20,22,24
- 1,2,3
Correct

Incorrect

Hint

a, 7, 23, c are in AP
$\begin{array}{l}7 - a = b - 7 = 23 - b = c - 23 = d\\b - 7 = 23 - b\\2b = 30\\b = 15\end{array}$
$\begin{array}{l}7 - a = b - 7\\ \Rightarrow 7 - a = 15 - 7 = 8\\ \Rightarrow a - 1\\23 - b = c - 23\\23 - 15 = c - 23\\8 = c - 23\\c = 31\end{array}$
Question 18 of 30

18. Question
1 point(s)
Find the 18th term of the AP whose 7th term is 24 less than the 11th term first term being 12?
- 120
- 114
- 130
- 150
Correct

Incorrect

Hint

$\begin{array}{l}a + 6d = a + 10d - 24\\ \Rightarrow 24 = 4d \Rightarrow d = 6\\{a_{18}} = a + 17d\\ = 12 + 17 \times 6\\ = 114\end{array}$
Question 19 of 30

19. Question
1 point(s)
The value of K so that${k^2} + 4k + 8$, $2{k^2} + 3k + 6$and $3{k^2} + 4k + 4$are three consecutive terms of an AP is?
- 0
- 1
- 2
- 3
Correct

Incorrect

Hint

$\begin{array}{l}(2{k^2} + 3k + 6) - ({k^2} + 4k + 8) = (3{k^2} + 4k + 4) - (2{k^2} + 3k + 6)\\ \Rightarrow 2{k^2} + 3k + 6 - {k^2} - 4k - 8 = 3{k^2} + 4k + 4 - 2{k^2} - 3k - 6\\ \Rightarrow {k^2} - k - 2 = {k^2} + k - 2\\ \Rightarrow - k = k\\ \Rightarrow 2k = 0\\ \Rightarrow k = 0\end{array}$
Question 20 of 30

20. Question
1 point(s)
The angles of a triangle are in AP the greatest angle is twice the least. Find all the angles of a triangle?
- $90^\circ ,45^\circ ,45^\circ $
- $60^\circ ,60^\circ ,60^\circ $
- $80^\circ ,60^\circ ,40^\circ $
- $70^\circ ,70^\circ ,40^\circ $
Correct

Incorrect

Hint

$\begin{array}{l}(2{k^2} + 3k + 6) - ({k^2} + 4k + 8) = (3{k^2} + 4k + 4) - (2{k^2} + 3k + 6)\\ \Rightarrow 2{k^2} + 3k + 6 - {k^2} - 4k - 8 = 3{k^2} + 4k + 4 - 2{k^2} - 3k - 6\\ \Rightarrow {k^2} - k - 2 = {k^2} + k - 2\\ \Rightarrow - k = k\\ \Rightarrow 2k = 0\\ \Rightarrow k = 0\end{array}$
Question 21 of 30

21. Question
1 point(s)
If sum of the 3rd and the 8th terms of an AP is 7 and sum of the 7th and 14th term is -3. The value of 8th term is?
- 0
- 1
- -1
- 3
Correct

Incorrect

Hint

$\begin{array}{l}{a_3} + {a_8} = 7\\{a_7} + {a_{14}} = - 3\\a + 2d + a + 7d = 7\\ \Rightarrow 2a + 9d = 7......(1)\\a + 6d + a + 13d = - 3\\ \Rightarrow 2a + 19d = - 3.....(2)\end{array}$
Substituting equation 1 from 2 we get,
$\begin{array}{l}10d = - 10 \Rightarrow d = - 1\\2a + 9d = 7\\ \Rightarrow 2a - 9 = 7 \Rightarrow a = 8\\{a_8} = a + 7d = 8 + 7 \times ( - 1) = 8 - 7 = 1\end{array}$
Question 22 of 30

22. Question
1 point(s)
How many numbers lie between 10 and 300 which divided by 4 leaves a remainder 3?
- 100
- 68
- 72
- 73
Correct

Incorrect

Hint

A=11, d=15-11=4
$\begin{array}{l}{a_n} = a + (n - 1)d\\299 = 11 + (n - 1)4\\288 = 4(n - 1)\\n - 1 = 72\\n = 73\end{array}$
Question 23 of 30

23. Question
1 point(s)
The sum of $( – 1) + ( – 2) + ( – 5) + …… + ( – 236)$is?
- 1000
- 9400
- -9400
- 10000
Correct

Incorrect

Hint

$\begin{array}{l}d = - 2 - 1 = - 3\\{S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]\\ = \dfrac{n}{2}\left[ {2 - 3n + 3} \right]\\ = \dfrac{n}{2}\left[ {5 - 3n} \right].....(1)\\n = a + (n - 1)d\\ - 236 = 1 + (n - 1)( - 3)\\ - 237 = - (n - 1) \times 3\\n - 1 = 79\\ \Rightarrow n = 80\\{S_n} = \dfrac{{80}}{2}\left[ {5 - 3 \times 80} \right]\\ = 40(5 - 240) = 40 \times ( - 235)\\ = - 9400\end{array}$
Question 24 of 30

24. Question
1 point(s)
The numbers 28, 22, x, y, 4 are in AP. The reciprocal values of x and y are?
- 10,16
- 20,18
- 18,16
- 16,10
Correct

Incorrect

Hint

$\begin{array}{l}22 - 28 = x - 22 = y - x = 4 - y\\ - 6 = x - 22\\x = - 6 + 22 = 16\\x - 22 = y - x\\ \Rightarrow 2x = y + 22\\ \Rightarrow 2 \times 16 = y + 22\\ \Rightarrow 32 - 22 = y\\ \Rightarrow y = 10\end{array}$
Question 25 of 30

25. Question
1 point(s)
The value of ‘n’ if the sum of the first ‘n’ terms of the AP 15, 23, 31,…. is 708
- 9
- 8
- 12
- 10
Correct

Incorrect

Hint

$\begin{array}{l}{S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]\\708 = \dfrac{n}{2}\left[ {30 + (n - 1)8} \right]\\1416 = n\left[ {30 + 8n - 8} \right]\\1416 = n\left[ {22 + 8n} \right]\\22n + 8{n^2} - 1416 = 0\\8{n^2} + 22n - 1416 = 0\\ \Rightarrow 4{n^2} + 11n - 708 = 0\\ \Rightarrow 4{n^2} + 59n - 48n - 708 = 0\\n(4n + 59) - 12(4n + 59) = 0\\(n - 12)(4n + 59) = 0\\n = 12,n = \dfrac{{ - 59}}{4}\end{array}$
But n cannot be negative so, n=12
Question 26 of 30

26. Question
1 point(s)
If the sum of p terms of an AP is q and the sum of q terms is p, then the sum of (p+q) terms will be.
- 0
- p-q
- P+q
- -(p+q)
Correct

Incorrect

Hint

Sum of p terms=q
$\begin{array}{l}{S_p} = \dfrac{p}{2}\left[ {2a + (p - 1)d} \right] = q\\ \Rightarrow p\left[ {2a + (p - 1)d} \right] = 2q\\2ap + p(p - 1)d = 2q.....(1)\end{array}$
Sum of q terms is p
$\begin{array}{l}{S_q} = \dfrac{q}{2}\left[ {2a + (q - 1)d} \right] = p\\q\left[ {2a + (q - 1)d} \right] = 2p\\2aq + q(q - 1)d = 2p.....(2)\end{array}$
Subtracting 2 from 1
$\begin{array}{l}2a(p - q) + \left[ {{p^2} - p - {q^2} + q} \right]d = 2q - 2p\\ \Rightarrow 2a(p - q) + \left[ {{p^2} - {q^2} - (p - q)} \right]d = - 2(p - q)\\ \Rightarrow 2a(p - q) + \left\{ {(p + q)(p - q)\_(p - q)} \right\}d = - 2(p - q)\\ \Rightarrow 2a(p - q) + (p - q)\left[ {p + q - 1} \right]d = - 2(p - q)\end{array}$
Dividing by p-q we get
$\begin{array}{l} = 2a + (p + q - 1)d = - 2\\{S_{p + q}} = \dfrac{{p + q}}{2}\left[ {2a + (p + q - 1)d} \right]\\ = \dfrac{{p + q}}{2} \times ( - 2) = - (p + q)\end{array}$
Question 27 of 30

27. Question
1 point(s)
If the sum of n terms of an AP is $3{n^2} + n$and its common difference is 6 then its first term is
- 2
- 4
- 3
- 7
Correct

Incorrect

Hint

$\begin{array}{l}{S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] = 3{n^2} + n\\ \Rightarrow \dfrac{n}{2} = \left[ {2a + (n - 1)6} \right] = 3{n^2} + n\\ \Rightarrow 2a + 6n - 6 = (3{n^2} + n) \times \dfrac{2}{n}\\ \Rightarrow 2a + 6n - 6 = n(3n + 1) \times \dfrac{2}{n}\\ \Rightarrow 2a + 6n - 6 = 6n + 2\\ \Rightarrow 2a = 8\\ \Rightarrow a = 4\end{array}$
Question 28 of 30

28. Question
1 point(s)
The first and last term of an AP is 1 and 11. If the sum of the term is 36, then the number of terms will be?
- 5
- 6
- 7
- 8
Correct

Incorrect

Hint

$\begin{array}{l}{a_n} = l = a + (n - 1)d = 11\\ \Rightarrow 1 + (n - 1)d = 11\\ \Rightarrow (n - 1)d = 10......(1)\\{S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] = 36\\ \Rightarrow \dfrac{n}{2}\left[ {2 \times 1 + 10} \right] = 36\\ \Rightarrow n\left[ {2 + 10} \right] = 72\\ \Rightarrow n = 6\end{array}$
Question 29 of 30

29. Question
1 point(s)
If the sum of n terms of an AP is $2{n^2} + 5n$then its nth term is?
- 4n-3
- 3n-4
- 4n+3
- 3n+4
Correct

Incorrect

Hint

$\begin{array}{l}{S_n} = 2{n^2} + 5n\\{S_1} = 2{(1)^2} = 5 = 7\\{S_2} = 2{(2)^2} + 10 = 18\\{a_1} = 7\\{a_2} = {S_2} - {S_1} = 11\\d = 11 - 7 = 4\\{a_n} = a + (n - 1)d\\ = 7 + (n - 1) \times 4\\ = 4n + 3\end{array}$
Question 30 of 30

30. Question
1 point(s)
If the sum of first n even natural number is equal to k times the sum of first n odd natural numbers, then k?
- $\dfrac{1}{n}$
- $\dfrac{{n – 1}}{n}$
- $\dfrac{{n + 1}}{{2n}}$
- $\dfrac{{n + 1}}{n}$
Correct

Incorrect

Hint

Sum of n even natural number = n(n+1)
Sum of n odd natural number = ${n^2}$
According to the question
$\begin{array}{l}n(n + 1) = k{n^2}\\ \Rightarrow k = \dfrac{{n(n + 1)}}{{{n^2}}} = \dfrac{{n + 1}}{n}\end{array}$