This set of Mathematics Online Test for Class 10 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

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

Explanation: Here a = 10, d = 2 and n = 20.

We know that, S

_{n}= \(\frac {n}{2}\) (2a + (n – 1)d)

S

_{20}= \(\frac {20}{2}\) (2(10) + (20 – 1)2)

S

_{20}= 10(20 + 38) = 580

3. The sum of 50 + 100 + 150 + 200 + …….. + 1000 is _________

a) 10000

b) 14400

c) 10500

d) 12100

View Answer

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

Explanation: n

^{th}term = 5n + 2

T

_{n}= 5n + 2

a = T

_{1}= 5(1) + 2 = 7

d = T

_{2}– T

_{1}= 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

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

6. The sum of all 2 – digit numbers divisible by 19 will be __________

a) 345

b) 245

c) 275

d) 285

View Answer

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

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} = 4n^{2} + 6, then its first term will be _________

a) 4

b) 3

c) 5

d) 1

View Answer

Explanation: The sum of first n terms of an AP is given by S

_{n}= 4n

^{2}+ 6.

Substituting n by n – 1 in the given equation

S

_{n – 1}= 4(n – 1)

^{2}+ 6

S

_{n – 1}= 4n

^{2}– 8n + 4 + 6

S

_{n – 1}= 4n

^{2}– 8n + 10

∴ T

_{n}= S

_{n}– S

_{n – 1}

T

_{n}= 4n

^{2}+ 6 – (4n

^{2}– 8n + 10)

T

_{n}= 8n – 4

The first term will be T

_{1}= 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

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 = 5n

^{2}+ 5n

n

^{2}+ 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

Explanation: The sum of the first 10 terms of an AP is the same as the sum of its first 5 terms

S

_{10}= S

_{5}

\(\frac {10}{2}\) (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

S

_{15}= \(\frac {n}{2}\) (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

Explanation: The sum of the first 12 terms of an AP is 15

S

_{12}= 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

S

_{15}= 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, S

_{27}= \(\frac {n}{2}\) (2a + (n – 1)d) = \(\frac {27}{2}\) (2a + (27 – 1)d) = \(\frac {27}{2}\) (2a + 26d) = 27(a + 13d) = -27 from (3).

**Sanfoundry Global Education & Learning Series – Mathematics – Class 10**.

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