This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Nth Term of Arithmetic Progression”.

1. What will be the n^{th} term of the AP 1, 5, 9, 13, 17…….?

a) 4n – 3

b) 3n – 4

c) 4n + 3

d) 3n + 4

View Answer

Explanation: Here a = 1 and d = 4

The n

^{th}term of the AP = a + (n – 1)d = 1 + (n – 1)4 = 1 + 4n – 4 = 4n – 3

2. If the n^{th} term of the AP is 2n + 7, then the common difference will be _____

a) -2

b) 2

c) 3

d) -3

View Answer

Explanation: n

^{th}term of the AP is 2n + 7

T

_{n}= 2n + 7

Common difference = T

_{2}– T

_{1}= (2 × 2 + 7 – (2 × 1 + 7)) = 11 – 9 = 2

3. If the n^{th} term of the AP is 7n – 9, then the first term is ________

a) 1

b) 6

c) 3

d) 4

View Answer

Explanation: n

^{th}term of the AP is 7n – 9

T

_{n}= 7n – 1

T

_{1}= 7(1) – 1 = 6

The first term of AP is 6.

4. If the n^{th} term of the AP is 8n + 1, then the 20^{th} term will be ______

a) 160

b) 120

c) 161

d) 121

View Answer

Explanation: n

^{th}term of the AP is 8n + 1

T

_{n}= 8n + 1

T

_{20}= 8(20) + 1

T

_{20}= 161

5. What will be the 99^{th} term from the end of the AP 500, 489, 478, 467… – 1139?

a) 1078

b) 1123

c) 12

d) 61

View Answer

Explanation: In this case since we have to find the 99

^{th}term from the end.

We will consider the first term to be -1139 and the common difference will be 11

Now, a = -1139, d = 11 and n = 99

T

_{99}= a + (n – 1)d

T

_{99}= -1139 + (99 – 1)11

T

_{99}= -1139 + 1078 = -61

The value of 99

^{th}term from the end is 61.

6. If the p^{th} term of an AP is q and its q^{th} term is p, then what will be the value of its (p + q)^{th} term?

a) 1

b) p + q – 1

c) 0

d) 2(p + q – 1)

View Answer

Explanation: p

^{th}term = q

a + (p – 1)d = q

a + pd – d = q (1)

q

^{th}term = p

a + (q – 1)d = p

a + qd – d = p (2)

Subtracting (2) from (1) we get,

a + qd – d – (a + pd – d) = p – q

qd – pd = p – q

d = -1

Substituting in equation 1, we get,

a = p + q – 1

(p + q)

^{th}term = a + (n – 1)d = p + q – 1 + (p + q – 1)(-1) = p + q – 1 – p – q + 1 = 0

7. If 5 times the 5^{th} term of an AP is equal to 15 times its 15^{th} term, then the value of its 20^{th} term will be _______

a) 0

b) 1

c) 2

d) 3

View Answer

Explanation: 5(5

^{th}term) = 15(15

^{th}term)

5

^{th}term = a + (5 – 1)d = a + 4d

15

^{th}term = a + (15 – 1)d = a + 14d

5(a + 4d) = 15(a + 14d)

5a + 20d = 15a + 210d

20d – 210d = 15a – 5a

-190d = 10a

a = -19d

Now, the 20

^{th}term = a + (20 – 1)d = a + 19d

But a = -19d

Hence, -19d + 19d = 0

8. If the 11^{th} term of an AP is \(\frac {1}{13}\) and its 13^{th} term is \(\frac {1}{11}\), then what will be the value of 143^{th} term?

a) \(\frac {1}{143}\)

b) 1

c) 0

d) \(\frac {23}{143}\)

View Answer

Explanation: Here 11

^{th}term = \(\frac {1}{13}\)

13

^{th}term = \(\frac {1}{11}\)

Let the first term of the AP be a and common difference be d

T

_{11}= a + (n – 1)d = \(\frac {1}{13}\)

T

_{11}= a + (11 – 1)d = \(\frac {1}{13}\)

T

_{11}= a + 10d = \(\frac {1}{13}\) (1)

T

_{13}= a + (n – 1)d = \(\frac {1}{11}\)

T

_{13}= a + (13 – 1)d = \(\frac {1}{11}\)

T

_{13}= a + 12d = \(\frac {1}{11}\) (2)

Subtracting (1) from (2)

We get,

a + 12d – (a + 10d) = \(\frac {1}{11} – \frac {1}{13}\)

2d = \(\frac {2}{143}\)

d = \(\frac {1}{143}\)

Now, substituting value of d in equation 1

We get,

T

_{11}= a + 10(\(\frac {1}{143}\)) = \(\frac {1}{13}\)

a = \(\frac {1}{143}\)

The 143

^{th}term = \(\frac {1}{143}\) + 142\(\frac {1}{143} = \frac {(1+142)}{143} \) = 1

9. If the 7^{th} term of an AP is 20 and its 11^{th} term is 40 then, what will be the common difference?

a) 3

b) 4

c) 5

d) 2

View Answer

Explanation: Here 7

^{th}term = 20

11

^{th}term = 40

Let the first term of the AP be a and common difference be d

T

_{7}= a + (n – 1)d = 20

T

_{7}= a + (7 – 1)d = 20

T

_{7}= a + 6d = 20 (1)

T

_{11}= a + (n – 1)d = 40

T

_{11}= a + (11 – 1)d = 40

T

_{11}= a + 10d = 40 (2)

Subtracting (1) from (2)

We get,

a + 10d – (a + 6d) = 40 – 20

4d = 20

d = 5

10. What will be the 14^{th} term of the AP 5, 8, 11, 14, 17…….?

a) 30

b) 41

c) 40

d) 44

View Answer

Explanation: Here a = 5, d = 8 – 5 = 3 and n = 14

T

_{14}= a + (n – 1)d

T

_{14}= 5 + (14 – 1)3

T

_{14}= 5 + 13 × 3

T

_{14}= 44

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

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