Class notes Mathametics and science

Date : 17-08-2025 STD 10 Maths(ARITHMETIC PROGRESSIONS) Total Marks : 30
Time : 1 Hour 30 Minute WEEKLY TEST 6


Section A

➤I Choose the right answer from the given options. [1 Marks Each] [10]
1. The first three terms of an AP respectively are 3y − 1, 3y + 5 and 5y + 1 . Then y equals:
(A) -3 (B) 4 (C) 5 (D) 2
Ans. : (c)
If and are in AP,
a, b c




b−a = c−b

2b = a + c

2(3y + 5) = 3y − 1 + 5y + 1

6y + 10 = 8y

10 = 8y − 6y

2y = 10

y = 5




Hence the correct option is (c).
2. An AP 5, 12, 19, ... has 50 terms. Its last term is
(A) 343 (B) 353 (C) 348 (D) 362
Ans. : (c)

Given, first term (a) = 5,

Common difference d = 12 − 5 = 7

th
∴ 50 term = a + (n − 1)d

= 5 + (50 − 1) × 7

= 5 + 49 × 7 = 348




3. How many two-digit numbers are divisible by 3 ?
(A) 25 (B) 30 (C) 32 (D) 36
Ans. : (b)
Two-digit numbers divisible by 3 are 12, 15, 18 , ...., 99 .
Let , Then
Tn = 99 12 + (n − 1) × 3 = 99




⇒ (n − 1)3 = 99 − 12

87
⇒ (n − 1) = = 29
3


⇒ n = 29 + 1 = 30




4. What is the 20 th term from the end of the AP 3, 8 , 13, ⋯ − 253 ?
(A) 163 (B) 158 (C) 153 (D) 148




Page 1

, Ans. : (b)
th
20 term from the end = [1 − (n − 1)d]


= (253 − 19 × 5)

= 253 − 95 = 158.




5. What is the common difference of an AP in which a 18 − a14 = 32 ?
(A) 8 (B) ` -8 (C) 4 (D) -4

Ans. : (a)

a18 − a14 = 32

a + 17d − a − 13d = 32

32
⇒ 4d = 32 ⇒ d = = 8
4




Hence, common difference is 8 .
6. If the n
th
term of an A.P. is (2n + 1) , then the sum of its first three terms is
(A) 6n + 3 (B) 15 (C) 12 (D) 21
Ans. : (b)
We have,
an = (2n + 1)

⇒ a1 = 2 × 1 + 1 = 3



So, the given sequence is an A.P. with first term a = a1 = 3 .
And the second term, . a2 = 2 × 2 + 1 = 5


So, the common difference,
d = a2 − a1 = 5 − 3 = 2

Therefore, the sum of first 3 terms of the A.P. is given by
n
Sn = {2a + (n − 1)d}
2

3
= {6 + (3 − 1)2}
2

3 3
= (6 + 4) = (10) = 15
2 2




Hence the correct option is (b).
7. If the sum of the first n terms of an A.P be 3n
2
+n and its common difference is 6 , then its first term is
(A) 2 (B) 3 (C) 1 (D) 4

Ans. : (d)
In the given problem, the sum of n terms of an A.P. is given by the expression,
2
Sn = 3n +n


Here, we can find the first term by substituting n = 1 as sum of first term of the A.P.
will be the same as the first term. So we get,

2
Sn = 3n +n
2
S1 = 3(1) + (1)

= 3+1

= 4




Therefore, the first term of this A.P is a = 4 .
8. The 13 term of an AP is 4 times its 3
th rd
term. If its 5
th
term is 16 then the sum of its first 10 terms is.
(A) 150 (B) 175 (C) 160 (D) 135




Page 2