Solutions of the above MCQs of Arithmetic Progression is given below
Solutions
Question 1; The 16th term of the A. P. 13, 7, 1, −5, −11, ——- is
- (a) − 77
- (b) 103
- (c) − 512
- (d) 928
Solutions: a = 13, d = 7 − 13 = − 6
tn = a + (n − 1 ) d
16th term = t16 = 13 + (16 −1) (−6)= 13 − 90 = − 77
| Option (a) |
Question 2: If (3/4), a, 2 are in A.P., then a =?
- (a) (11/8)
- (b) (7/8)
- (c) (3/8)
- (d) (5/8)
Solution: As (3/4), a, 2 are in A.P.,
a − (3/4) = 2 − a
⇒ 2a = 2 + (3/4)
⇒ 2a = (11/4)
⇒ a = (11/8)
| Option (a) |
Question 3: Which term of the A.P. 4, 9, 14, 19, …….. is 79?
- (a) 15th
- (b) 16th
- (c) 14th
- (d) 18th
Solution: a = 4, d = 9 − 4 = 5 , let nth term be 79
As tn = a + (n − 1) d
⇒ 79 = 4 + (n −1) 5
⇒ 79 = 4 + 5 n − 5
⇒ 5 n = 80 ⇒ n = 16
| Option (b) |
Question 4: The nth term of an A.P. is given by Tn = (4n − 7)
(i) Find its first term
- (a) 11
- (b) − 3
- (c) 1
- (d) 4
Solution: Tn = (4 n −7)
∴ first term = T1 = 4 × 1 − 7 = − 3
| Option (b) |
(ii) Find common difference
- (a) 11
- (b) −3
- (c) 1
- (d) 4
Solution: T2 = 4 × 2 − 7 = 1
T3 = 4 × 3 − 7 = 5
T4 = 4 × 4 − 7 = 9
∴ common difference = T2 − T1 = T3 − T2 = 4
| Option (d) |
(iii) Find 2nd term
- (a) 11
- (b) −3
- (c) 1
- (d) 4
Solution: 2nd term = T2 = 1
| Option (c) |
Question 5: Which term of the A.P. 92, 88, 84, …… is 0?
- (a) 23rd
- (b) 24th
- (c) 27th
- (d) 28th
Solution: a = 92, d = 88 − 92 = − 4
Let nth term be 0. So tn = 0
tn = a + (n −1) d
⇒ 0 = 92 +(n −1) (−4)
⇒ 0 = 92 − 4n + 4
⇒ 4n = 96 ⇒ n = 24
| Option (b) |
Question 6: Which term of the A.P. 41, 38, 35, ….. is the first negative term?
- (a) 14th
- (b) 15th
- (c) 16th
- (d) 17th
Solution: a = 41, d = − 3
Let nth term be the first negative term.
∴ tn < o
⇒ 41 + (n −1) (−3) < 0
⇒ 41 − 3 n + 3 < 0
⇒ 44 < 3 n
⇒ 3 n > 44
⇒ n > (44/3)
⇒ n > 14 + (2/3)
So first negative term is 15th term
| Option (b) |
Question 7: The 10th term from the end of an A.P. 7, 10, 13, …., 184 is
- (a) 151
- (b) 154
- (c) 160
- (d) 157
Solution: The series from the end is: 184, ……, 13, 10, 7
So a = 184, d = 10 − 13 = − 3
10th term = t10 = 184 + (10 −1) (−3)
= 184 − 27 = 157
| Option (d) |
Question 8: How many two-digit numbers are there
which are divisible by 6?
- (a) 14
- (b) 15
- (c) 16
- (d) 17
Solution: The series is 6, 12, 18, …….., 96
a = 12, d = 6, let tn = 96
as tn = a + (n −1) d
⇒ 96 = 12 + (n − 1) 6
⇒ 96 = 6 n + 6
⇒ 6 n = 90 ⇒ n = 15
| Option (b) |
Question 9: 5 + 9 + 13 + 17 + …….. up to 23 terms = ?
- (a) 1123
- (b) 1125
- (c) 1127
- (d) none of these
Solution: a = 5, d = 9 − 5 = 4 , n = 23
S23 = (n/2) {2a + (n −1) d}
= (23/2) {2 × 5 + (23 −1) 4
= (23/2) (10 + 88)
= (23/2) × 98 = 1127
| Option (c) |
Question 10: How many terms of the A.P. 6, 12, 18, 24, …… must
be taken to make the sum 816?
- (a) 14
- (b) 16
- (c) 17
- (d) 20
Solution: a = 6, d = 12 − 6 = 6 , let Sn = 816
As Sn = (n/2) {2a + (n −1) d}
⇒ 816 = (n/2) {2 × 6 + (n −1) 6}
⇒ 816 = (n/2) (12 + 6n −6)
⇒ 816 = (n/2) (6n +6)
⇒ 816 = 3n(n+1)
⇒ 272 = n² + n
⇒ n² + n − 272 = 0
⇒ n² + 17 n – 16 n – 272 = 0
⇒ (n + 17) (n − 16) = 0
either n = −17 (no of terms can’t be negative), n = 16
| Option (b) |
Question 11: The angles of a triangle are in A.P.
whose common difference is thirty degrees. Find the angles.
- (a) 40°, 60°, 80°
- (b) 30°, 60°, 90°
- (c) 40°, 70°, 100°
- (d) 60°, 90°, 120°
Solution: d = 30°, let the angles are a, a+ 30°, a+60°
∴ a + a + 30° + a + 60° = 180°
⇒ 3a = 90° ⇒ a = 30°
the angles of the triangle are 30°, 60°, 90°
| Option ((b) |
Question 12: In a flower bed there are 32 rose plants in the
first row, 30 in the second row, 28 in the third row, and so on.
There are 10 rose plants in the last row.
(i) How many rows are there in the flower bed?
- (a) 10
- (b) 11
- (c) 12
- (d) 15
Solution: The series is 32, 30, 28, ……, 10
a = 32, d = 30 − 32 = −2, let no of rows = n. So tn = 10
tn = a + (n −1) d
⇒ 10 = 32 + (n −1) (−2)
⇒ 10 = 32 − 2n + 2
⇒ 2n = 24 ⇒ n = 12
| Option (c) |
(ii) How many rose plants are there in the flower bed?
- (a) 242
- (b) 252
- (c) 280
- (d) 516
Solution: Total no of rose plants
= S12 = (12/2) {2 × 32 + (12 −1) (−2)
= 6 (64 −22) = 252
| Option (b) |
Question 13: Find the sum of the first 15 multiples of 8.
- (a) 540
- (b) 900
- (c) 952.5
- (d) 960
Solution: The series is 8, 16, 24, 32, …. up to 15 term
a = 8, d = 16 −8 = 8, n = 15
∴ S15 = (15/2) { 2 × 8 + (15 −1) 8}
= (15/2) (16 + 112)
= (15/2) × 128 = 960
| Option (d) |
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