Sum of N Terms of Arithmetic Progression - Class 10 Maths MCQ

This set of Class 10 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Sum of N Terms of Arithmetic Progression”.

1. The sum of n terms of an AP in which first term is a, common difference is d and last term is l, is S_n = \(\frac {n}{2}\)(a + l).
a) True
b) False
View Answer

Answer: a
Explanation: Consider an AP having n terms in which
First term = a, common difference = d and last term = l.
Then, l = a + (n – 1)d
We may write the given AP as a, a + d, a + 2d …., (l – 2d), (l – d), l
Let S_n be the sum of the first n terms of the AP. Then,
S_n = a + (a + d) + (a + 2d) … + (l – 2d) + (l – d) + l
Writing the above series in reverse order, we get
S_n = (l – 2d) + (l – d) + l … + a + (a + d) + (a + 2d)
Adding the two equations we get,
2S_n = a + l + (a + l) + (a + l) … n times = n(a + l)
S_n = \(\frac {n}{2}\)(a + l)

2. The sum of first 20 terms of the AP 10, 12, 14, 16, 18…. is _________
a) 200
b) 580
c) 620
d) 440
View Answer

Answer: b
Explanation: Here a = 10, d = 2 and n = 20.
We know that, S_n = \(\frac {n}{2}\) (2a + (n – 1)d)
S₂₀ = \(\frac {20}{2}\) (2(10) + (20 – 1)2)
S₂₀ = 10(20 + 38) = 580

3. The sum of 50 + 100 + 150 + 200 + …….. + 1000 is _________
a) 10000
b) 14400
c) 10500
d) 12100
View Answer

Answer: c
Explanation: Here a = 50, d = 50 and T_n = 1000.
T_n = a + (n – 1)d = 1000
50 + (n – 1)50 = 1000
50 + 50n – 50 = 1000
n = 20
We know that, S_n = \(\frac {n}{2}\) (a + l)
S_n = \(\frac {20}{2}\) ((50) + 1000)
S_n = 10(50 + 1000) = 10500

4. The n^th term if an AP is 5n + 2, then the sum of first n terms of the AP will be __________
a) \(\frac {n^2 – 8}{2}\)
b) \(\frac {n^2 + 8}{2}\)
c) \(\frac {n^2 – 9}{2}\)
d) \(\frac {n^2 + 9}{2}\)
View Answer

Answer: d
Explanation: n^th term = 5n + 2
T_n = 5n + 2
a = T₁ = 5(1) + 2 = 7
d = T₂ – T₁ = 5(2) + 2 – (5 + 2) = 5
We know that, S_n = \(\frac {n}{2}\) (2a + (n – 1)d)
S_n = \(\frac {n}{2}\) (2(7) + (n – 1)5)
S_n = 7n + \(\frac {n^2}{2} – \frac {5n}{2} = \frac {n^2 + 9}{2}\)

5. What will be the sum of all 3-digit even positive numbers?
a) 247050
b) 269450
c) 269350
d) 269580
View Answer

Answer: a
Explanation: The AP according to the given data will be 100, 102, 104, …, 998
Here a = 100 and d = 2
T_n = a + (n – 1)d = 998
100 + (n – 1)2 = 998
100 + 2n – 2 = 998
2n = 998 – 98
n = 450
We know that, S_n = \(\frac {n}{2}\) (a + l)
S_n = \(\frac {450}{2}\) ((100) + 998)
S_n = 225(100 + 998) = 247050

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6. The sum of all 2 – digit numbers divisible by 19 will be __________
a) 345
b) 245
c) 275
d) 285
View Answer

Answer: c
Explanation: The AP according to the given data will be 19, 38, 57, …, 95
Here a = 19 and d = 19
T_n = a + (n – 1)d = 95
19 + (n – 1)19 = 95
19 + 19n – 19 = 95
19n = 95
n = 5
We know that, S_n = \(\frac {n}{2}\) (a + l)
S_n = \(\frac {5}{2}\) ((19) + 95)
S_n = 285

7. What will be the sum of all natural numbers lying between 500 and 5000, which are divisible by 17?
a) 567898
b) 729810
c) 765835
d) 234526
View Answer

Answer: b
Explanation: The AP according to the given data will be 510, 527, 544, …, 4998
Here a = 510 and d = 17
T_n = a + (n – 1)d = 4998
510 + (n – 1)17 = 4998
510 + 17n – 17 = 4998
17n = 4505
n = 265
We know that, S_n = \(\frac {n}{2}\) (a + l)
S_n = \(\frac {265}{2}\) ((510) + 4998)
S_n = 132.5(5508) = 729810

8. If the sum of first n terms of an AP is given by S_n = 4n² + 6, then its first term will be _________
a) 4
b) 3
c) 5
d) 1
View Answer

Answer: a
Explanation: The sum of first n terms of an AP is given by S_n = 4n² + 6.
Substituting n by n – 1 in the given equation
S_{n – 1} = 4(n – 1)² + 6
S_{n – 1} = 4n² – 8n + 4 + 6
S_{n – 1} = 4n² – 8n + 10
∴ T_n = S_n – S_{n – 1}
T_n = 4n² + 6 – (4n² – 8n + 10)
T_n = 8n – 4
The first term will be T₁ = 8(1) – 4 = 4

9. How many terms of the AP 5,10,15,20….. must be added to get a sum of 330?
a) -11
b) -12
c) 12
d) 11
View Answer

Answer: d
Explanation: Here a = 5, d = 5 and S_n = 330
We know that, S_n = \(\frac {n}{2}\) (2a + (n – 1)d)
330 = \(\frac {n}{2}\) (2(5) + (n – 1)5)
330 = \(\frac {n}{2}\) (10 + 5n – 5)
330 = \(\frac {n}{2}\) (5 + 5n)
330 = \(\frac {5n^2+5n}{2}\)
660 = 5n² + 5n
n² + n – 132 = 0
n = -12, 11
Since, n cannot be negative
Hence value of n is 11.

10. The sum of the first 10 terms of an AP is the same as the sum of its first 5 terms, then the sum of its 15th term will be _______
a) 1
b) 0
c) 2
d) 3
View Answer

Answer: b
Explanation: The sum of the first 10 terms of an AP is the same as the sum of its first 5 terms
S₁₀ = S₅
\(\frac {10}{2}\) (2a + (10 – 1)d) = \(\frac {5}{2}\) (2a + (5 – 1)d)
5(2a + 9d) = \(\frac {5}{2}\) (2a + 4d)
5(2a + 9d) = 5(a + 2d)
(2a + 9d) = (a + 2d)
2a – a = 2d – 9d
a = – 7d
S₁₅ = \(\frac {n}{2}\) (2a + (n – 1)d)
= \(\frac {15}{2}\) (2a + (15 – 1)d)
= \(\frac {15}{2}\) (2a + 14d)
= 15(a + 7d)
= 15(-7d + 7d) = 0

11. The sum of the first 12 terms of an AP is 15 and the sum of its first 15 terms is 12, then the sum of its first 27 terms will be __________
a) -24
b) 24
c) 27
d) -27
View Answer

Answer: d
Explanation: The sum of the first 12 terms of an AP is 15
S₁₂ = 15
\(\frac {n}{2}\) (2a + (n – 1)d) = 15
\(\frac {12}{2}\) (2a + (12 – 1)d) = 15
6(2a + 11d) = 15
12a + 66d = 15
4a + 22d = 5 (1)
The sum of its first 15 terms is 12
S₁₅ = 12
\(\frac {n}{2}\) (2a + (n – 1)d) = 12
\(\frac {15}{2}\) (2a + (15 – 1)d) = 12
\(\frac {15}{2}\) (2a + 14d) = 12
15(a + 7d) = 12
15a + 105d = 12
5a + 35d = 4 (2)
On subtracting (2) from (1)
4a + 22d – (5a + 35d) = 5 – 4
-a – 13d = 1
a + 13d = -1 (3)
Now, S₂₇ = \(\frac {n}{2}\) (2a + (n – 1)d) = \(\frac {27}{2}\) (2a + (27 – 1)d) = \(\frac {27}{2}\) (2a + 26d) = 27(a + 13d) = -27 from (3).

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