Arithmetic Progression Class 10 MCQ CBSE Maths

Arithmetic Progression Class 10 MCQ with Solution

1. Find the 20^th term of the given Arithmetic Progression 5, 8, 11, . . . .

(A) 60

(B) 62

(D) 61

Ans. (B)

Sol. T₂₀ = 5 + (20 – 1) 3

T₂₀ = 5 + 57 = 62

2. Ajay started work in 1995 at an annual salary of Rs.5000 and received an increment of Rs.200 each year. In which year did his income reach Rs.7000 –

(A) 19^th

(B) 11^th

(D) 15^th

Ans. (B)

Sol. Change in income = Rs. 7000 – Rs. 5000

So time = $\frac{{2000}}{{20}} = 10$ years

i.e. In the 11^th year he will receive Rs. 7000

3. In an A.P. , the sum of first n terms is $\frac{{3{n^2}}}{2}\, + \,\frac{{5n}}{2}$ . Find its 25^th term.

(A) 74

(B) 75

(D) 77

Ans. (C)

Sol. Now t_n = S_n– S_n–1

t_n = $\left( {\frac{{3{n^2}}}{2} + \frac{{5n}}{2}} \right) - \,\left( {\frac{{3{{(n - 1)}^2}}}{2} + \frac{{5(n - 1)}}{2}} \right)$

= $\frac{3}{2}{n^2} + \frac{5}{2}n - \left[ {\frac{3}{2}({n^2} + 1 - 2n) + \frac{5}{2}(n) - \frac{5}{2}} \right]$

= $\frac{3}{2}{n^2} + \frac{5}{2}n - \frac{3}{2}{n^2} - \frac{3}{2} + 3n - \frac{5}{2}n + \frac{5}{2}$

= $\frac{5}{2} - \frac{3}{2} + 3n\,\, \Rightarrow \,\,\frac{2}{2} + 3n\,\, \Rightarrow \,\,1 + 3n$

Now n = 25

the 25^th term is 1 + 3 × 25 = 76

4. If sum of n terms in an AP is given S_n = 2n² then find the 20^th term :

(A) 80

(B) 78

(D) None of these

Ans. (B)

Sol. T_n = S_n – S_n–1

= 2n² – [2(n – 1)² ]

= 4n – 2

So T₂₀ = 78

5. If a, b, c are in A.P. and a – c = b then find the value of a in terms of b :

(A) $\frac{{2b}}{3}$

(B) $\frac{{3b}}{2}$

(D) $\frac{{b + 2}}{2}$

Ans. (B)

Sol. a – c = b (given)

a + c = 2b ( a, b, c are in A.P.)

Adding 2a = 3b

∴ a = $\frac{{3b}}{2}$

6. Find the number of terms in A.P. If sum of these terms is 710 and first and last terms are 7 and 64 respectively.:

(A) 19

(B) 20

(D) 10

Ans. (B)

Sol. Sum = 710 = (7 + 64) K/2 ⇒ K = 20

7. If first, second last and last term of an A.P. are –6, 26 and 30 respectively. Then find the number of terms:

(A) 10

(B) 11

(D) 8

Ans. (A)

Sol. Given a = –6, t_n–1 = 26, & t_n = 30

t_n – t_n _{– 1} = d = 30 – 26 = 4

t_n = a +(n – 1)d ⇒ 30 = –6 + (n – 1) 4

n – 1 = $\frac{{36}}{4}$ = 9

So n = 10

8. Which term of the sequence 17, 25, 33, 41….. is just greater than 90.:

(A) 11

(B) 12

(D) 13

Ans. (A)

Sol. Let t_n> 90 ⇒ t_n = 17 + (n – 1) 8 > 90

⇒ 9 + 8n > 90 ⇒ n > $\frac{{81}}{8}$

So n = 11

9. If a, A₁, A₂, b are in A.P. then find $\frac{{{A_1}}}{{{A_2}}}$ :

(A) $\frac{{2a\, + \,b}}{{a + b}}$

(B) $\frac{{2a\,\, + \,\,b}}{{a\,\, + \,\,2b}}$

(D) $\frac{{2a\,\, + \,3b}}{{a\,\, + \,\,b}}$

Ans. (B)

Sol. We know d = $\frac{{b - a}}{{n + 1}}$ and n = 2 So d = $\frac{{b - a}}{3}$ $\frac{{{A_1}}}{{{A_2}}}$ = $\frac{{a + \left( {\frac{{b - a}}{3}} \right)}}{{a + 2\left( {\frac{{b - a}}{3}} \right)}} = \frac{{2a + b}}{{a + 2b}}$

10. If the product of roots of equation (a – b) x²+ (b – c) x + (c – a) = 0 is 1 then :

(A) 2a = b + c

(B) 2b = a + c

(D) b = 2a + c

Ans. (A)

Sol. Given product = 1

⇒ $\frac{{c - a}}{{a - b}} = 1$

⇒ 2a = b + c

11. Fifth term of an AP is 8. What is the sum of the first 9 terms of the AP :

(A) 56

(B) 72

(D) None of these

Ans. (B)

Sol. T₅ = a + 4d = 8

s₉ = $\frac{9}{2}\,[2a + (9\, - \,1)d\,]$

= $\frac{9}{2} \times \,2[a\, + \,4d]$

= 9 × 8

= 72

12. The sum of 10 terms of the series $\sqrt 3 \, + \,\sqrt {12} \, + \,\sqrt {27} \, + \,\sqrt {48} \, + \,....$ is :

(A) 55

(B) 66

(D) $55\sqrt 3$

Ans. (D)

Sol. $\sqrt 3 \, + \sqrt {12} \, + \,\sqrt {27} \, + \,\sqrt {48} \,.....$

⇒ $\sqrt 3 \, + \sqrt {12} \, + \,\sqrt {27} \, + \,\sqrt {48} \,.....$

s₁₀ = $\frac{{10}}{2}\left[ {2 \times \,\sqrt 3 \, + \,(10\, - \,1)\, \times \,\sqrt 3 } \right]$

= $55\sqrt 3$

13. The interior angles of a convex polygon are in A.P. If the smallest angle be 120° and the common difference be 5°, then the number of sides is :

(A) 9

(B) 16

(D) 6

Ans. (A)

Sol. a = 120°

d = 5°

(2n –4)×90 = $\frac{n}{2}\,[2 \times \,120\, + \,(n\, - \,1) \times \,5]$

On solving

n =9 or 16

but n =16 polygon will be concave

so n = 9

14. If p^th term of an AP is $\frac{1}{q}$ and q^th term is $\frac{1}{p}$ , then the sum of first pq terms is :

(A) pq

(B) 0

(D) $\frac{1}{2}$ (pq+1)

Ans. (D)

Sol. t_p = a + (p–1) d = $\frac{1}{q}$ ….(1)

t_q = a + (q–1)d = $\frac{1}{p}$ ….(2)

(1) – (2) (p – q)d = $\frac{{p-q}}{{pq}}$

∴ d = $\frac{1}{{pq}}$

putting d in (1) a = $\frac{1}{q}-\frac{{p-1}}{{pq}} = \frac{1}{{pq}}$

∴ s_pq = $\frac{{pq}}{2}\left[ {2x\frac{1}{{pq}} + (pq-1) \times \frac{1}{{pq}}} \right] = \frac{{pq}}{2}\left[ {\frac{2}{{pq}} + 1-\frac{1}{{pq}}} \right]$ = $\frac{{pq}}{2}\left[ {\frac{{pq + 1}}{{pq}}} \right]$

= $\frac{1}{2}[pq + 1]$

15. If sum of m terms of an AP is n and sum of n terms is m, then the sum of (m+n) terms is :

(A) mn

(B) 0

(D) –(m+n)

Ans. (D)

Sol. s_m = $\frac{m}{2}\,\left[ {2a\, + (m\, - \,1)\,d} \right]\, = \,n$ …(1)

s_n = $\frac{n}{2}\,\left[ {2a\, + (n\, - \,1)\,d} \right]\, = \,m$ …(2)

s_m+n= $\frac{{m\, + \,n}}{2}\,[2a + (m + n - 1)d]$ …(3)

From (1) and (2) we will find a & d and will put in (3)

∴ s_m+n = –(m+n)

16. If (k+1), 3k and (4k+2) be any three consecutive terms of an AP, then the value of K is :

(A) 3

(B) 0

(D) 2

Ans. (A)

Sol. 2 × 3k = k + 1+ 4k + 2

⇒ k = 3

17. The sum of all 2 digit numbers is ;

(A) 4750

(B) 4895

(D) 4680

Ans. (C)

Sol. s = 10 + 11 + 12 + ……99 = $\frac{{90}}{2}(10 + 99) = 45 \times 109$

= 4905

18. The 6th and 8th terms of an AP are 12 and 22 respectively. Its 2nd term is :

(A) 8

(B) –8

(D) –3

Ans. (B)

Sol. T₆ = a + 5d = 12 ….(1)

T₈ = a + 7d = 22 ….(2)

From (1) & (2) a = – 13, d = 5

so, T₂ = a + d

= – 13 + 5

= – 8

19. If the sum of n terms of an A.P. is 2n² + 5n , then its n^th term is :

(A) 4n – 3

(B) 3n – 4

(D) 3n + 4

Ans. (C)

Sol. s_n = 2n² + 5n

s_n–1 = 2(n –1)² + 5(n–1)

T_n = s_n –s_n–1

= 2n² + 5n – [2(n–1)² + 5(n–1)]

T_n = 4n + 3

20. If the sum of first n even natural numbers is equal to k times the sum of first n odd natural number, then K =:

(A) $\frac{{\rm{1}}}{{\rm{n}}}$

(B) $\frac{{n - 1}}{n}$

(D) $\frac{{n + 1}}{n}$

Ans. (C)

Sol. s_even = 2 + 4 + 6 + ……2n = $\frac{n}{2}\,[2 \times \,2\, + \,(n\, - 1)\, \times 2]\, = \,n(n + 1)$

s_odd= 1 + 3 + 5 + …….2n –1 = $\frac{n}{2}\,[2 \times 1\, + \,(n\, - 1)\, \times 2]\, = \,{n^2}$

so s_even = k.s_odd

n(n+1) = k.n²

k = $\frac{{n + 1}}{n}$

21. If 7th and 13th terms of an A.P be 34 and 64 respectively, than its 18th term is :

(A) 87

(B) 88

(D) 90

Ans. (C)

Sol. T₇ = a + 6d = 34 …… (1)

T₁₃= a + 12d = 64 …… (2)

from (1) & (2) a = 4

d = 5

so, T₁₈= a + 17d

= 4 + 17 × 5

= 89

22. Sum of n terms of the series $\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} +$ ….. is :

(A) $\frac{{n(n + 1)}}{2}$

(B) 2n (n+1)

(D) 1

Ans. (C)

Sol. s_n = $\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} +$

s_n = $\sqrt 2 \, + 2\sqrt 2 \, + \,3\sqrt 2 \, + \,.....$

s_n = $\frac{{\rm{n}}}{{\rm{2}}}[2 \times \,\sqrt 2 \, + \,(n\, - \,1)\,\sqrt 2 ]$

= $\frac{{n(n\, + \,1)}}{{\sqrt 2 }}$

23. If the A.M. of ‘a’ and ‘b’ is $\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}$ , then the value of n is :

(A) 1

(B) $\frac{1}{2}$

(D) 0

Ans. (D)

Sol. $\frac{{a\, + \,b}}{2}$ = $\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}$

From this n = 0

24. The value of $\sum\limits_{k = 1}^{15} {(2k - 3)}$ is :

(A) 390

(B) 195

(D) 426

Ans. (B)

Sol. s = $\sum\limits_{k = 1}^{15} {(2k - 3)}$

s = –1 + 1 + 3 + 5 + …….27

s = 195

25. The sum of the first 100 positive integers exactly divisible by 7 is :

(A) 35350

(B) 35700

(D) 1393

Ans. (A)

Sol. s₁₀₀ = 7 + 14 + 21 + ….

= $\frac{{100}}{2}\left[ {2 \times \,7\, + \,(100\, - \,1) \times \,7} \right]$

= 50[14 + 693]

= 50 × 707

= 35350

26. The sum of 20 terms of log 2 + log 4 + log 8 + …….. is :

(A) 20 log 2

(B) Log 20

(D) log 2

Ans. (C)

Sol. s₂₀ = log 2 + log 4 + log 8 + ……..

s₂₀ = log 2 + 2 log 2 + 3 log 2 + ……..

s₂₀ = 210 log 2

27. Which term of the AP, 24, 21, 18, 15 …….. is the first negative :

(A) 8

(B) 10

(D) 6

Ans. (B)

Sol. a = 24

d = –3

so let T_n is first negative term

t_n = a + (n–1) d < 0

24 + (n –1) × (–3) < 0

24 < (n–1) × 3

8 < (n–1)

9 < n

so, n = 10

28. The ratio of the sum of m and n terms of an A.P. is m² : n² . Then the ratio of the m^th and n^th terms is :

(A) $\frac{{2n - 1}}{{2n + 1}}$

(B) $\frac{{2m - 1}}{{2n - 1}}$

(D) $\frac{{2m + 1}}{{2m - 1}}$

Ans. (B)

Sol. $\frac{{{s_m}}}{{{s_n}}}$ = $\frac{{{m^2}}}{{{n^2}}}$ $\frac{{\frac{m}{2}\left[ {2a\, + \,(m\, - \,1)\,d} \right]}}{{\frac{n}{2}\,[2a + (n - 1)\,d]}}$ = $\frac{{{m^2}}}{{{n^2}}}$ $\frac{{m\left[ {2a\, + \,(m\, - \,1)\,d} \right]}}{{n[2a + (n - 1)\,d]}}$ = $\frac{{{m^2}}}{{{n^2}}}$ $\frac{{\left[ {2a\, + \,(m\, - \,1)\,d} \right]}}{{2a\, + \,(n\, - \,1)\,d}}$ = $\frac{m}{n}$

Putting m = 2m –1

n = 2n –1

so $\frac{{2a\, + \,(2m\, - 1 - 1)d}}{{2a + (2m - 1 - 1)d}}$ = $\frac{{2m\, - 1}}{{2n\, - \,1}}$ $\frac{{2[a + (m - 1)\,d]}}{{2[a + (n - 1)\,d]}}$ = $\frac{{2m\, - 1}}{{2n\, - \,1}}$ $\frac{{{T_m}}}{{{T_n}}}$ = $\frac{{2m\, - 1}}{{2n\, - \,1}}$

29. If n A.M. are inserted between 20 and 80 such that the ratio of first to last mean is 1 : 3, The value of n is :

(A) 11

(B) 15

(B) 17

(D) 5

Ans. (A)

Sol. 20, a, a₂ ……….a_n ,₈₀

so there are n + 2 term

t_n+2 = a + (n+2 –1) d = 80

20 + (n+1) d = 80

d = $\frac{{60}}{{n + 1}}$

so t_n+1 = a_n = a+(n+1 –1) d = a + nd

= 20 + n × $\frac{{60}}{{4 + 1}}$

T₂ = a₁ = a + d

= 20 + $\frac{{60}}{{n + 1}}$

so $\frac{{{a_n}}}{{{a_1}}}$ = $\frac{1}{3}$ (given)

\frac{{20\, + n \times \,\frac{{60}}{{n + 1}}}}{{20 + \frac{{60}}{{n + 1}}}}

\frac{3}{1}

so n = 11

30. If there are (2n + 1) terms in A.P. then the ratio of the sum of odd terms and the sum of even terms is :

(A) n + 1 : n – 1

(B) n + 1 : n²

(D) None of these

Ans. (C)

Sol. Let A.P. = a, a+d, a + 2d , ……

so $\frac{{{s_{even}}}}{{{s_{odd}}}}$ = $\frac{{\frac{n}{2}\left[ {2(a + d)\, + \,(n\, - \,1)2d} \right]}}{{\frac{{n + 1}}{2}\,\left[ {2a\, + \,(n\, + 1\, - \,1)\,2d} \right]}}$

= $\frac{{\frac{n}{2}[2a\, + \,2d\, + \,2nd\, - \,2d]}}{{\frac{{n\, + \,1}}{2}\,[2a\, + \,2nd]}}$ $\frac{{{s_{even}}}}{{{s_{odd}}}}$ = $\frac{{n(2a\, + \,2nd)}}{{n + 1(2a\, + \,2nd)}}$ $\frac{{{s_{even}}}}{{{s_{odd}}}}$ = $\frac{n}{{n\, + \,1}}$

so $\frac{{{s_{odd}}}}{{{s_{even}}}}$ = $\frac{{n + 1}}{{n\,}}$

31. Fourth term of an arithmetic progression is 8. What is the sum of the first 7 terms of the arithmetic progression:

(A) 7

(B) 64

(D) Can’t be determined

Ans. (C)

Sol. T₄ = a + 3d =8

S₇ = $\frac{7}{2}\,[2a\, + \,(7\, - 1)\,d]$

= $\frac{7}{2}\, \times \,[2a\, + \,6d]$

= $\frac{7}{2} \times \,2\,[a\, + 3\,d]$

= $\frac{7}{2} \times \,2\, \times \,8$ = 56

32. If the sides of a right angled triangle are in A.P. then the sides are proportional to:

(A) 1 : 2 : 3

(B) 2 : 3 : 4

(D) 4 : 5 : 6

Ans. (C)

Sol. a² + b² = c² …..(1)

2b = a + c

so form (1) and (2)

a : b : c = 3 : 4 : 5

33. If a clock strikes once at a one o’clock, twice at two o’ clock and twelve times at 12 o’clock and again once at one o’clock and so on, how many times will the bell be struck in the course of 2 days :

(A) 156

(B) 108

(D) None of these

Ans. (D)

Sol. s = 2 × 2 [1+ 2 + 3 + ….. 12]

= 4 × $\frac{{12 \times 13}}{2}$ = 4 × 78

= 312

34. If the sum of the series 2 + 5 + 8 + 11……. is 60100, then the number of terms are :

(A) 100

(B) 200

(D) 250

Ans. (B)

Sol. s_n = 2 + 5 + 8 + 11 + …..

60100 = $\frac{n}{2}[2 \times 2 + (n--1) \times 3]$

so n = 200

35. 8^th term of series $2\sqrt 2$ , $\sqrt 2$ , 0, …..will be:

(A) $- 5\sqrt 2$

(B) $5\sqrt 2$

(D) $- 10\sqrt 2$

Ans. (A)

Sol. a = $2\sqrt 2$

d = $\sqrt 2 \, - \,2\sqrt 2 \,\, = \,\, - \sqrt 2$

T₈ = a + 7d = $2\sqrt 2 \, + \,7\, \times \,( - \sqrt 2 )$ = $5\sqrt 2$

36. 30^th term of the A.P. 10, 7, 4, …… is :

(A) 97

(B) 77

(D) –-87

Ans. (C)

Sol. T₃₀ = 10 + (30 –1) × (–3)

T₃₀ = –77

37. 11^th term of A.P. –3, $- \,\frac{1}{2}$ , 2 …… is

(A) 28

(B) 22

(D) – $48\frac{1}{2}$

Ans. (B)

Sol. T₁₁ = a + (11–1) d

= $- 3\, + \,10\left( {\frac{5}{2}} \right)$

T₁₁ = 22

38. If the sum of the series 54 + 51 + 48 + …… is 513, then the number of terms are –

(A) 18

(B) 20

(D) None of these

Ans. (A)

Sol. S_n = 54 + 51 + 48 + ……..

513 = $\frac{n}{2}$ [2×54 + (n–1) × (–3) ]

so n = 18

39. There are 60 terms in an A.P. of which the first term is 8 and the last term is 185. The 31^st term is :

(A) 56

(B) 94

(D) 248

Ans. (D)

Sol. a = 8

T₆₀ = 185

T₆₀ = a + (60–1)d

185 = 8 + 59.d

177 = 59 d

d = 3

T₃₁ = a + (31 –1) d

= 8 + 30 × 8 = 248

40. If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are :

(A) 5, 10, 15, 20

(B) 4, 10, 16, 22

(D) None of these

Ans. a –3d+a –d+a+d+a +3d = 50

4a = 50

a = 12.5

a +3d = 4(a–3d)

On solving d = 2.5

so number are 5, 10, 15, 20

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