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## Answers

2nd term is 3 and 1st term is 1 so 3-1=2, common difference=2

to find n's value using formula a+(n-1)d=249(last term)

putting the value of a and d : 1+(n-1)2=249

1+2n-2=249

-1+2n=249

2n=249+1

2n=250

n=125

now sum of series=n/2[a+L]

putting the value of n,a,L

then 125/2[1+249]

125/2[250]

125*125=15625

*given,*

1+3+5+7+............249

here the given list of numbers are in a.p.so the sum of n numbers of an a.p is

sn=n/2(a+l)...............(1)

here,

sn is the sum of the given list of numbers,

n is the no. of numbers in the given list,

a is the 1st number in the list,

l is the last number in the list.

here,to find n

we know that

an=a+(n-1)d

249=1+(n-1)2

249=1+2n-2

n=125

sub this value in (1)

sn=n/2(a+l)

=125/2(1+249)

=125/2x250

=15625

hope this helped u.......:))

1+3+5+7+............249

here the given list of numbers are in a.p.so the sum of n numbers of an a.p is

sn=n/2(a+l)...............(1)

here,

sn is the sum of the given list of numbers,

n is the no. of numbers in the given list,

a is the 1st number in the list,

l is the last number in the list.

here,to find n

we know that

an=a+(n-1)d

249=1+(n-1)2

249=1+2n-2

n=125

sub this value in (1)

sn=n/2(a+l)

=125/2(1+249)

=125/2x250

=15625

hope this helped u.......:))