The sum of even numbers between 1 and 31 is
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REQUIRED SUM=(2+4+6+8+........+30)
(2+4+6+8+.............+2n)=n(n+1)
here, 2n=30
n=30/2=15
15(15+1)=15*16=240 hope this helps !
(2+4+6+8+.............+2n)=n(n+1)
here, 2n=30
n=30/2=15
15(15+1)=15*16=240 hope this helps !
sank258otge54:
thanks pretty
S₁₅ = n/2 [ a + l ]
Answered by
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Hi friend !
The first set of even nos: between 1 and 31 are :-
2 , 4 , 6 , .............. and the last one is 30
2 , 4 , 6 ....... 30
here ,
first term = a = 2
common difference = d = 2
last term = l = 30
no: of even no:s between 1 and 31 = n
we know that ,
l = a + [ n - 1 ] d
30 = 2 + [ n - 1 ] 2
30 - 2 = ( n - 1) 2
28 /2 = n - 1
14 = n - 1
n = 15
we have a formula to calculate the sum ,
S₁₅ = n/2 [ a + l ]
= 15/2 x [ 2 + 30 ]
= 15/2 x 32
= 240
=============================
Sum of even no:s between 1 and 31 is 240
The first set of even nos: between 1 and 31 are :-
2 , 4 , 6 , .............. and the last one is 30
2 , 4 , 6 ....... 30
here ,
first term = a = 2
common difference = d = 2
last term = l = 30
no: of even no:s between 1 and 31 = n
we know that ,
l = a + [ n - 1 ] d
30 = 2 + [ n - 1 ] 2
30 - 2 = ( n - 1) 2
28 /2 = n - 1
14 = n - 1
n = 15
we have a formula to calculate the sum ,
S₁₅ = n/2 [ a + l ]
= 15/2 x [ 2 + 30 ]
= 15/2 x 32
= 240
=============================
Sum of even no:s between 1 and 31 is 240
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