Sequences and Series
- Only two entries are known of the following Arithmetic progression : ––, 5, ––, ––, 14, ––, --What should be the number just after 14 ?
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For an arithmetic sequence,
tn = a + (n – 1) d
∴ 5 = a + (2 – 1) d
⇒ 5 = a + d .........(i)
and 14 = a + 4d ......(ii)
By subtracting equation (i) from (ii),
∴ d = 9 = 3 3
Correct Option: A
For an arithmetic sequence,
tn = a + (n – 1) d
∴ 5 = a + (2 – 1) d
⇒ 5 = a + d .........(i)
and 14 = a + 4d ......(ii)
By subtracting equation (i) from (ii),∴ d = 9 = 3 3
From equation (i),
5 = a + 3 ⇒ a = 5 – 3 = 2
∴ t6 = 2 + (6 – 1) × 3
t6 = 2 + 15 = 17
- Which term of the sequence 7, 10, 13, ..... is 151 ?
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Let the nth term = 151
Here, first term = a = 7
common difference = d = 3
∴ tn = a + (n – 1) d
⇒ 151 = 7 + (n – 1) × 3
⇒ (n – 1) × 3 = 144Correct Option: D
Let the nth term = 151
Here, first term = a = 7
common difference = d = 3
∴ tn = a + (n – 1) d
⇒ 151 = 7 + (n – 1) × 3
⇒ (n – 1) × 3 = 144⇒ n – 1 = 144 = 48 3
⇒ n = 49
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The sum of the first 20 terms of the series 1 + 1 + 1 + ..... is 5 × 6 6 × 7 7 × 8
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From the question ,
First term = 1 = 1 - 1 5 × 6 5 6 Second term = 1 = 1 - 1 6 × 7 6 7 20th term = 1 = 1 - 1 24 × 25 24 25 ∴ Required sum = 1 - 1 + 1 - 1 + ...... + 1 - 1 5 6 6 7 24 25
Correct Option: A
From the question ,
First term = 1 = 1 - 1 5 × 6 5 6 Second term = 1 = 1 - 1 6 × 7 6 7 20th term = 1 = 1 - 1 24 × 25 24 25 ∴ Required sum = 1 - 1 + 1 - 1 + ...... + 1 - 1 5 6 6 7 24 25 Required sum = 1 - 1 = 5 - 1 = 4 5 25 25 25
Required sum = 0.16
- Which term of the sequence 6, 13, 20, 27, ..... is 98 more than its 24th term ?
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The 24th term of the sequence 6 ,13 , 20 , 27........
Here , a = 6 , n = 24 , d = 7
∴ Tn = a + (n – 1) d
∴ T24 = 6 + (24 – 1) 7
T24 = 6 + 23 × 7 = 6 + 161 = 167
Let the required nth term = 265
∴ 265 = 6 + ( n – 1) 7
⇒ (n – 1) 7 = 265 – 6 = 259Correct Option: B
The 24th term of the sequence 6 ,13 , 20 , 27........
Here , a = 6 , n = 24 , d = 7
∴ Tn = a + (n – 1) d
∴ T24 = 6 + (24 – 1) 7
T24 = 6 + 23 × 7 = 6 + 161 = 167
Let the required nth term = 265
∴ 265 = 6 + ( n – 1) 7
⇒ (n – 1) 7 = 265 – 6 = 259⇒ n – 1 = 259 = 37 7
⇒ n – 1 = 37 ⇒ n = 38
- The sum of series 1 + 2 + 3 + 4 + .... + 998 + 999 + 1000 is
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We know that , 1 + 2 + 3 + 4 + ...... + n = n(n + 1) 2 ∴ 1 + 2 + 3 + 4 + ...... + 1000 = 1000(1000 + 1) 2
Correct Option: B
We know that , 1 + 2 + 3 + 4 + ...... + n = n(n + 1) 2 ∴ 1 + 2 + 3 + 4 + ...... + 1000 = 1000(1000 + 1) 2 sum = 1000 × 1001 = 500500 2
