Sequences and Series
- The sum of the first 8 terms of a geometric progression is 6560 and the common ratio is 3. The first term is
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As we know that ,
Sum of x terms of a GP = a(rn - 1) (when r > 1) r - 1
Here , Sum = 6560 , n = 8 , r = 3∴ 6560 = a(38 - 1) 3 - 1
Correct Option: B
As we know that ,
Sum of x terms of a GP = a(rn - 1) (when r > 1) r - 1
Here , Sum = 6560 , n = 8 , r = 3∴ 6560 = a(38 - 1) 3 - 1 ⇒ a = 6560 × 2 ⇒ a = 2 6560
- If the 4th term of an arithmetic progression is 14 and the 12th term is 70, then the first term is :
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Here , a4 = 14 , a12 = 70
Using the given formula ,
a4 = a1 + (4 – 1) × d
14 = a1 + 3d
⇒ a1 = 14 – 3d ....(i)
70 = a1 + 11d ....(ii)
After putting the value of a1 in equation (i)
⇒ 14 – 3d + 11d = 70Correct Option: B
Here , a4 = 14 , a12 = 70
Using the given formula ,
a4 = a1 + (4 – 1) × d
14 = a1 + 3d
⇒ a1 = 14 – 3d ....(i)
70 = a1 + 11d ....(ii)
After putting the value of a1 in equation (i)
⇒ 14 – 3d + 11d = 70
⇒ 8d = 70 – 14
⇒ d = 7
∴ a1 = 14 – 21 = – 7
- The sum 9 + 16 + 25 + 36 + .... + 100 is equal to :
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As per the given question ,
Sum = 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 380
Second method to find the required answer with the help of formula :
Sn = 9 + 16 + 25 + ......... + 100
Sn = 3² + 4² + 5² + .......... + 10²
Sn = (1² + 2² + 3² + 4² +......+10²) – 1² – 2²Sn = n(n + 1)(2n + 1) - 5 6
Correct Option: B
As per the given question ,
Sum = 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 380
Second method to find the required answer with the help of formula :
Sn = 9 + 16 + 25 + ......... + 100
Sn = 3² + 4² + 5² + .......... + 10²
Sn = (1² + 2² + 3² + 4² +......+10²) – 1² – 2²Sn = n(n + 1)(2n + 1) - 5 6 Sn = 10(10 + 1)(2 × 10 + 1) - 5 6 Sn = 10 × 11 × 21 - 5 6
Sn = 55 × 7 – 5 = 385 – 5 = 380
- The sum (101 + 102 + 103 + .... + 200) is equal to :
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Let S = 101 + 102 + 103 + .... + 200
S = (100 + 1) + (100 + 2) + (100 + 3) + ...+ (100 + 100)
Thus, it consists of 100 terms.
S = (100 + 100 + 100 + .... 100 times) + (1 + 2 + 3 + ...... + 100)
S = (100 × 100) + (1 + 2 + 3 + ..... + 100)
S = (10000) + (1 + 2 + 3 + ... + 100)S = 10000 + 100 × (100 + 1) 2
S = 10000 + 5050 = 15050
We can find the required answer with the help of given formula :
Here, a = 101, d = 102 – 101 = 1 , L = 200
∴ L = an = a + (n – 1)d
200 = 101 + (n – 1)1
⇒ n – 1 = 99
n = 100Sn = n [a + L] 2
Correct Option: C
Let S = 101 + 102 + 103 + .... + 200
S = (100 + 1) + (100 + 2) + (100 + 3) + ...+ (100 + 100)
Thus, it consists of 100 terms.
S = (100 + 100 + 100 + .... 100 times) + (1 + 2 + 3 + ...... + 100)
S = (100 × 100) + (1 + 2 + 3 + ..... + 100)
S = (10000) + (1 + 2 + 3 + ... + 100)S = 10000 + 100 × (100 + 1) 2
S = 10000 + 5050 = 15050
We can find the required answer with the help of given formula :
Here, a = 101, d = 102 – 101 = 1 , L = 200
∴ L = an = a + (n – 1)d
200 = 101 + (n – 1)1
⇒ n – 1 = 99
n = 100Sn = n [a + L] 2 Sn = 100 [101 + 200] 2
∴ Sn = 50 × 301 = 15050
- Find the wrong number in the following number series.
3 , 7 , 16 , 35 , 70 , 153
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The pattern is :
3 × 2 + 1 = 6 + 1 = 7
7 × 2 + 2 = 14 + 2 = 16
16 × 2 + 3 = 32 + 3 = 35
........... and so on.Correct Option: A
The pattern is :
3 × 2 + 1 = 6 + 1 = 7
7 × 2 + 2 = 14 + 2 = 16
16 × 2 + 3 = 32 + 3 = 35
35 × 2 + 4 = 70 + 4 = 74 ≠ 70
Here , 70 is wrong term . So , 74 will be correct term .
74 × 2 + 5 = 148 + 5 = 153
Thus , required answer is 70.
