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Find the sum of all the two digit natural no. which is divisible by 4

Answer 1

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Raghunath answered 3 year(s) ago

Find the sum of all the two-digit natural number which are divisible by 4.

Find the sum of all the two-digit natural number which are divisible by 4. 

Class-X Maths

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Asked by David

Feb 19

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Raghunath , SubjectMatterExpert

Member since Apr 11 2014

The two digits numbers which are divisible by 4 are
12,16,20,......96  in AP.

Sn = n/2(2a+(n -1)d)

Here a = 12 , d = 4.

tn = a + (n - 1) d

96 = 12 + ( n - 1) 4

84 = ( n - 1) 4

n - 1 = 21

∴n = 22

S22 = (22/2) (2×12 + (22 -1)4)

S22 = 11×[ 24 + 84]

S22 = 11×[ 24 + 84]

∴S22 = 1188.

Answer 2

Hi ,

12 , 16 , 20 ,...96 are two digit

numbers which are divisible by 4

are in A.P

first term = a = 12

Common difference = a2 - a1

d = 16 - 12 = 4

Last term = l = 96

a + ( n - 1 )d = 96

12 + ( n - 1 ) 4 = 96

( n - 1 ) 4 = 96 - 12

( n - 1 ) 4 = 84

n - 1 = 84/4

n = 21 + 1

n = 22

Therefore ,

Sum of 22 terms = S22

S22 = 22/2 [ 12 + 96 ]

= 11 × 108

= 1188

I hope this helps you.

: )