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Find the sum
31+33+35+37+.......+133
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The series forms an Arithmetic progression where,
a=30 and d=2
The sum of the first 'n' terms of an A.P. is given by the formula:
S=(n/2)*(a+l), where l=last term
'n' can be calculated using:
nth term=a+(n-1)d
thus, 133=31+(n-1)*2
On solving we get n=52
On substituting in the original formula, we get:
S=(52/2)*(31+133)
=26*164
=4264
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a=31
d=2
an=133
an=a+(n-1)d
133=31+(n-1)2
133=31+2n-2
133=29+2n
133-29=2n
104=2n
n=104/2
n=52
Sn=n/2(a+an)
=52/2(31+133)
=26(164)
=4,264