Number System
- The sum of all natural numbers from 75 to 97 is :
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Series of all natural numbers from 75 to 97 is in A.P. whose first term, a = 75, last term, l = 97
If number of terms be n, then
an = a + ( n – 1 )d
⇒ 97 = 75 + ( n – 1 )
⇒ n = 97 – 74 = 23Sn = n (a + 1) 2 Correct Option: D
Series of all natural numbers from 75 to 97 is in A.P. whose first term, a = 75, last term, l = 97
If number of terms be n, then
an = a + ( n – 1 )d
⇒ 97 = 75 + ( n – 1 )
⇒ n = 97 – 74 = 23Sn = n (a + 1) 2 S23 = 23 (75 + 97) 2 S23 = 23 × 172 = 1978 2
- The sum of first 20 odd natural numbers is equal to :
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Series of first 20 odd natural numbers is an arithmetic progression with 1 as the first term and the common difference 2. Sum of n terms in arithmetic progression is given by.
Sn = n [2a + (n − 1)d] 2
Where a : First term, d : common difference and n = Number of termsCorrect Option: C
Series of first 20 odd natural numbers is an arithmetic progression with 1 as the first term and the common difference 2. Sum of n terms in arithmetic progression is given by.
Sn = n [2a + (n − 1)d] 2
Where a = First term, d = common difference and n = Number of terms∴ S20 = 20 × [(2 × 1) + (20 − 1) × 2] 2
∴ S20 = 10 [ 2 + 38 ] = 10 × 40 = 400
Note : Sum of first n consecutive odd numbers = n2
- The sum of three consecutive odd natural numbers is 147. Then, the middle number is :
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Let us assume the 3 consecutive odd natural numbers are P, P + 2, P + 4;
According to question;
∴ P + P + 2 + P + 4 = 147
⇒ 3P + 6 = 147
⇒ 3P = 147 – 6 = 141Correct Option: C
Let us assume the 3 consecutive odd natural numbers are P, P + 2, P + 4;
According to question;
∴ P + P + 2 + P + 4 = 147
⇒ 3P + 6 = 147
⇒ 3P = 147 – 6 = 141⇒ P = 141 = 47 3
∴ Middle Number = P + 2 = 47 + 2 = 49
- The unit digit in 3 × 38 × 537 × 1256 is
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Unit’s digit in 3 × 38 × 537 × 1256
= Unit’s digit in 3 × 8 × 7 × 6Correct Option: D
Unit’s digit in 3 × 38 × 537 × 1256
= Unit’s digit in 3 × 8 × 7 × 6
= 4 × 2 = 8
- The digit in the unit’s place of the product (2464)1793 × (615)317 × (131)491 is
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(4)2m gives 6 at unit digit.
(4)2m +1 gives 4 at unit digit.
(5)n gives 5 at unit place.Correct Option: A
(4)2m gives 6 at unit digit.
(4)2m +1 gives 4 at unit digit.
(5)n gives 5 at unit place.
The same is the case with 1.
∴ Required digit = Unit’s digit in the product of 4× 5 × 1 = 0
