Mathematics Questions and Answers – Nth Term of Arithmetic Progression

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

1. What will be the nth term of the AP 1, 5, 9, 13, 17…….?
a) 4n – 3
b) 3n – 4
c) 4n + 3
d) 3n + 4
View Answer

Answer: a
Explanation: Here a = 1 and d = 4
The nth term of the AP = a + (n – 1)d = 1 + (n – 1)4 = 1 + 4n – 4 = 4n – 3
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2. If the nth term of the AP is 2n + 7, then the common difference will be _____
a) -2
b) 2
c) 3
d) -3
View Answer

Answer: b
Explanation: nth term of the AP is 2n + 7
Tn = 2n + 7
Common difference = T2 – T1 = (2 × 2 + 7 – (2 × 1 + 7)) = 11 – 9 = 2

3. If the nth term of the AP is 7n – 9, then the first term is ________
a) 1
b) 6
c) 3
d) 4
View Answer

Answer: b
Explanation: nth term of the AP is 7n – 9
Tn = 7n – 1
T1 = 7(1) – 1 = 6
The first term of AP is 6.
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4. If the nth term of the AP is 8n + 1, then the 20th term will be ______
a) 160
b) 120
c) 161
d) 121
View Answer

Answer: c
Explanation: nth term of the AP is 8n + 1
Tn = 8n + 1
T20 = 8(20) + 1
T20 = 161

5. What will be the 99th term from the end of the AP 500, 489, 478, 467… – 1139?
a) 1078
b) 1123
c) 12
d) 61
View Answer

Answer: d
Explanation: In this case since we have to find the 99th 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
T99 = a + (n – 1)d
T99 = -1139 + (99 – 1)11
T99 = -1139 + 1078 = -61
The value of 99th term from the end is 61.
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6. If the pth term of an AP is q and its qth 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

Answer: c
Explanation: pth term = q
a + (p – 1)d = q
a + pd – d = q     (1)
qth 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 5th term of an AP is equal to 15 times its 15th term, then the value of its 20th term will be _______
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: a
Explanation: 5(5th term) = 15(15th term)
5th term = a + (5 – 1)d = a + 4d
15th 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 20th term = a + (20 – 1)d = a + 19d
But a = -19d
Hence, -19d + 19d = 0
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8. If the 11th term of an AP is \(\frac {1}{13}\) and its 13th term is \(\frac {1}{11}\), then what will be the value of 143th term?
a) \(\frac {1}{143}\)
b) 1
c) 0
d) \(\frac {23}{143}\)
View Answer

Answer: b
Explanation: Here 11th term = \(\frac {1}{13}\)
13thterm = \(\frac {1}{11}\)
Let the first term of the AP be a and common difference be d
T11 = a + (n – 1)d = \(\frac {1}{13}\)
T11 = a + (11 – 1)d = \(\frac {1}{13}\)
T11 = a + 10d = \(\frac {1}{13}\)     (1)
T13 = a + (n – 1)d = \(\frac {1}{11}\)
T13 = a + (13 – 1)d = \(\frac {1}{11}\)
T13 = 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,
T11 = a + 10(\(\frac {1}{143}\)) = \(\frac {1}{13}\)
a = \(\frac {1}{143}\)
The 143th term = \(\frac {1}{143}\) + 142\(\frac {1}{143} = \frac {(1+142)}{143} \) = 1

9. If the 7th term of an AP is 20 and its 11th term is 40 then, what will be the common difference?
a) 3
b) 4
c) 5
d) 2
View Answer

Answer: c
Explanation: Here 7th term = 20
11thterm = 40
Let the first term of the AP be a and common difference be d
T7 = a + (n – 1)d = 20
T7 = a + (7 – 1)d = 20
T7 = a + 6d = 20     (1)
T11 = a + (n – 1)d = 40
T11 = a + (11 – 1)d = 40
T11 = a + 10d = 40     (2)
Subtracting (1) from (2)
We get,
a + 10d – (a + 6d) = 40 – 20
4d = 20
d = 5
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10. What will be the 14th term of the AP 5, 8, 11, 14, 17…….?
a) 30
b) 41
c) 40
d) 44
View Answer

Answer: d
Explanation: Here a = 5, d = 8 – 5 = 3 and n = 14
T14 = a + (n – 1)d
T14 = 5 + (14 – 1)3
T14 = 5 + 13 × 3
T14 = 44

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

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter