Math, asked by abdulshahal66, 7 months ago

Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 76 -

MCQ WORKSHEET-III

CLASS X: CHAPTER – 5

ARITHMETIC PROGRESSION

1. 7th term of an AP is 40. The sum of its first 13th terms is

(a) 500 (b) 510 (c) 520 (d) 530

2. The sum of the first four terms of an AP is 28 and sum of the first eight terms of the same AP is

88. Sum of first 16 terms of the AP is

(a) 346 (b) 340 (c) 304 (d) 268

3. Which term of the AP 4, 9, 14, 19, ….. is 109?

(a) 14th (b) 18th (c) 22nd (d) 16th

4. How many terms are there in the arithmetic series 1 + 3 + 5 + …….. + 73 + 75?

(a) 28 (b) 30 (c) 36 (d) 38

5. 51 + 52 + 53 + 54 +……. + 100 = ?

(a) 3775 (b) 4025 (c) 4275 (d) 5050

6. How many natural numbers between 1 and 1000 are divisible by 5?

(a) 197 (b) 198 (c) 199 (d) 200

7. If a, a – 2 and 3a are in AP, then the value of a is

(a) –3 (b) –2 (c) 3 (d) 2

8. How many terms are there in the AP 7, 10, 13, …. , 151?

(a) 50 (b) 55 (c) 45 (d) 49

9. The 4th term of an AP is 14 and its 12th term is 70. What is its first term?

(a) –10 (b) –7 (c) 7 (d) 10

10. The first term of an AP is 6 and the common difference is 5. What will be its 11th term?

(a) 56 (b) 41 (c) 46 (d) none of these

11. Which term of the AP 72, 63, 54, ……. is 0?

(a) 8th (b) 9th (c) 11th (d) 12th

12. The 8th term of an AP is 17 and its 14th term is –29. The common difference of the AP is

(a) –2 (b) 3 (c) 2 (d) 5

13. Which term of the AP 2, –1, –4, –7, ……. is –40?

(a) 8th (b) 15th (c) 11th (d) 23rd

14. Which term of the AP 20, 17, 14,………… is the first negative term?

(a) 8th (b) 6th (c) 9th (d) 7th

15. The first, second and last terms of an AP are respectively 4, 7 and 31. How many terms are there

in the given AP?

(a) 10 (b) 12 (c) 8 (d) 13

Answers

Answered by Swarup1998

Let the first term be a and the common difference be d.

Then 7th term is a + 6d.

Given, a + 6d = 40 ..... (1)

Now the sum of the first 13 terms of the A. P. is

= 13/2 * [2a + (13 - 1) * d]

= 13/2 * [2a + 12d]

= 13/2 * 2 [a + 6d]

= 13 * [a + 6d]

= 13 * 40, by (1)

= 520

Let the first term be a and the common difference be d.

Then the sum of the first 4 terms of the A. P. is

= 4/2 * [2a + (4 - 1) * d]

= 2 * [2a + 3d]

and the sum of the first 8 terms of the A. P. is

= 8/2 * [2a + (8 - 1) * d]

= 4 * [2a + 7d]

Given,

2 * [2a + 3d] = 28 or, 2a + 3d = 14 ..... (1)

4 * [2a + 7d] = 88 or, 2a + 7d = 22 ..... (2)

On subtraction, we get

4d = 8 or, d = 2

Putting d = 2 in (1), we get

2a + 3 * 2 = 14

or, 2a + 6 = 14

or, 2a = 8

or, a = 4

Now the sum of the first 16 terms of the A. P. is

= 16/2 * [2a + (16 - 1) * d]

= 8 * [2 * 4 + 15 * 2]

= 8 * [8 + 30]

= 8 * 38

= 304