CLASS 10TH , MCQ , CHEPTER 5 , ARITHMETIC PROGRESSION (WITH SOLUTION) , CBSE , PSEB & ALL BOARD EXAM

 

Important mcq ,class 10th ,chapter 5, Arithmetic Progression, mathematics ( answers given in the end)

1)

The nth term of an A.P. is given by an = 3 + 4n. The common difference is

(a) 7. (b) 3. (c) 4. (d) 1

2)

. If p, q, r and s are in A.P. then r – q is

(a) s – p. (b) s – q

(c) s – r. (d) none of these

3)  If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are

(a) 2, 4, 6.      (b) 1, 5, 3

(c) 2, 8, 4.      (d) 2, 3, 4

4) The (n – 1)th term of an A.P. is given by 7,12,17, 22,… is

(a) 5n + 2.              (b) 5n + 3

(c) 5n – 5.               (d) 5n – 3

5)     The 10th term from the end of the A.P. -5, -10, -15,…, -1000 is

(a) -955. (b) -945

(c) -950. (d) -965

6) The sum of all two digit odd numbers is

(a) 2575.                (b) 2475

(c) 2524.                 (d) 2425

7)  The sum of first n odd natural numbers is

(a) 2n².                  (b) 2n + 1

(c) 2n – 1.               (d) n²

8). If a, b, c, d, e are in A.P., then the value of a – 4b + 6c – 4d + e is

(a) 0.        (b) 1.     (c) -1.        (d) 2

9). The 10th term from the end of the A.P. 4, 9,14, …, 254 is

(a) 209. (b) 205. (c) 214. (d) 213

10) If 2x, x + 10, 3x + 2 are in A.P., then x is equal to

(a) 0. (b) 2. (c) 4. (d) 6

///////////////////  ANSWERS.  ///////////////////

1

Answer: c

Explaination:Reason: We have An = 3 + 4n

∴ An+1= 3 + 4(n + 1) = 7 + 4n

∴ d = An+1 – An

= (7 + 4n) – (3 + 4n)

= 7 – 3

= 4

2 .Answer: c

Explaination:Reason: Since p, q, r, s are in A.P.

∴ (q – p) = (r – q) = (s – r) = d (common difference)

3. Answer: d

Explaination:Reason: Let three numbers be a – d, a, a + d

∴ a – d +a + a + d = 9

⇒ 3a = 9

⇒ a = 3

Also (a – d) . a . (a + d) = 24

⇒ (3 -d) .3(3 + d) = 24

⇒ 9 – d² = 8

⇒ d² = 9 – 8 = 1

∴ d = ± 1

Hence numbers are 2, 3, 4 or 4, 3, 2

4 Answer: d

Explaination:Reason: Here a = 7, d = 12-7 = 5

∴ an-1 = a + [(n – 1) – l]d

 = 7 + [(n – 1) -1] (5)

 = 7 + (n – 2)5 = 7 + 5n – 10 = 5M – 3

5 Answer: a

Explaination:Reason: Here l = -1000, d = -10 – (-5) = -10 + 5 = – 5

∴ 10th term from the end = l – (n – 1 )d = -1000 – (10 – 1) (-5) = -1000 + 45 = -955

6 Answer: b

Explaination:Reason: All two digit odd numbers are 11,13,15,… 99, which are in A.P.

Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even

∴ Sum = 45/2[11 + 99] = 45/2 × 110 = 45 × 55 = 2475

7 Answer: d

Explaination:Reason: Required Sum = 1 + 3 + 5 + … + upto n terms.

Here a = 1, d = 3 – 1 = 2

Sum = n/2[2 × 1 + (n – 1) × 2] = n/2[2 + 2n – 2] = n/2 × 2n = n²

8 Answer: a

Explaination:Reason: Let common difference of A.P. be x

∴ b = a + x, c = a + 2x, d = a + 3x and e = a + 4x

Given equation n-4b + 6c-4d + c

= a – 4(a + x) + 6(a + 2x) – 4(a+ 3x) + (a + 4x)

= a – 4a – 4x + 6a + 12x – 4a – 12x + a + 4x = 8a – 8a + 16x – 16x = 0

9 Answer: a

Explaination:Reason: Here l – 254, d = 9-4 = 5

∴ 10th term from the end = l – (10 – 1 )d 

= 254 -9d = 254 - 9(5) = 254 – 45 = 209

10 Answer: d

Explaination:Reason: Since 2x, x + 10 and 3x + 2 are in A.P.

∴ 2(x + 10) = 2x + (3x + 2)

⇒ 2x + 20 – 5x + 2

⇒ 2x – 5x = 2 – 20

⇒ 3x = 18

⇒ x = 6