the sum when even nos. 2+4+6+8+10..........250 is
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Answered by
3
1st Method:
Sn (Sum of the 'n' numbers) = n/2 (2a+(n-1)d)
there given last no. is 250, so the no.of even numbers here is 250/2 i.e., 125
i.e., n=125, a (1st number) = 2, l (last number) = 250
then,
Sn = n/2 (a+l)
= 125/2 (2+250)
= 125×252/2
= 125(126)
= 15750
therefore, the sum of all the even no.s upto 250 is 15750
2nd Method:
Sn = n/2 (2a+(n-1)d)
we got n= 250/2 = 125, a=2, d (common difference) = 2
Sn = 125/2 (2(2)+(125-1)2)
= 125(4+248)/2
= 125(252)/2
= 125(126)
= 15750
hence, the sum of all the even no.s upto 250 is 15750
Sn (Sum of the 'n' numbers) = n/2 (2a+(n-1)d)
there given last no. is 250, so the no.of even numbers here is 250/2 i.e., 125
i.e., n=125, a (1st number) = 2, l (last number) = 250
then,
Sn = n/2 (a+l)
= 125/2 (2+250)
= 125×252/2
= 125(126)
= 15750
therefore, the sum of all the even no.s upto 250 is 15750
2nd Method:
Sn = n/2 (2a+(n-1)d)
we got n= 250/2 = 125, a=2, d (common difference) = 2
Sn = 125/2 (2(2)+(125-1)2)
= 125(4+248)/2
= 125(252)/2
= 125(126)
= 15750
hence, the sum of all the even no.s upto 250 is 15750
tabishabbasi786:
ri8 answer by A.P
wlcom
Answered by
1
This is an ap with common diff.. 2
so use formula n(a1+an)/2
on this ap there will be 125 terms
so ..n=125
therefore
Sn=125(2+250)/2
=(125*252)/2
=15750
so use formula n(a1+an)/2
on this ap there will be 125 terms
so ..n=125
therefore
Sn=125(2+250)/2
=(125*252)/2
=15750
plz mark as d best if it helps u
nth term = first term + (number of terms - 1)*common difference /2
applying it in the above example , nth term = 250 , first term = 2 and common difference = 2 , substitute these values and you will find the number of terms in Arithmetic Progression ..
and Finally calculate the sum of series = (first term + last term )*number of terms / 2 !!!
enjoy !!
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