If the sum of five consecutive even integers is t, then, in terms of t, what is the greatest integer?
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see the attachment dear!!
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Hi dude!
Here is the answer :
Assume the first integer as 'a'. As the sequence is formed by even integers the common difference d=2. The number of terms n=5 (given).
Sum of n terms=n/2(2a+(n-1)d)
Sum of 5 terms = 5/2(2a + (5-1)(2))
=5/2(2a + 8)
= 5(a+4)=5a+20=t
Therefore, 5a=t-20
a=(t-20)/5
'n' th term of A.P = a+(n-1)d
'5'th of the A.P=a+(5-1)(2)
=(t-20)/5+(4)(2)
=(t-20)/5+8
=(t-20)/5+40/5
=(t-20+40)/5
=(t+20)/5
Hope your are satisfied my answer,dude.
Here is the answer :
Assume the first integer as 'a'. As the sequence is formed by even integers the common difference d=2. The number of terms n=5 (given).
Sum of n terms=n/2(2a+(n-1)d)
Sum of 5 terms = 5/2(2a + (5-1)(2))
=5/2(2a + 8)
= 5(a+4)=5a+20=t
Therefore, 5a=t-20
a=(t-20)/5
'n' th term of A.P = a+(n-1)d
'5'th of the A.P=a+(5-1)(2)
=(t-20)/5+(4)(2)
=(t-20)/5+8
=(t-20)/5+40/5
=(t-20+40)/5
=(t+20)/5
Hope your are satisfied my answer,dude.
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