Find the sum of first 20 terms of A.P whose nth term is 4n-1.
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given that
nth term is 4n-1
then
if n=1 ; 4(1)-1 = 3
n= 2 ; 4(2)-1 = 7
n= 3 ; 4(3)-1 = 11
----------------------------------
their common difference is 4
so that these are in the A.p.
3,7,11,...........................
the sum of nth term formula is
sn = n/2[2a+(n-1)d]
the sum of first 20 terms are
= 20/2[2(3)+(20-1)4]
= 10[6+(19)4]
= 10×[6+76]
= 10(82)
= 820.
the sum of the first 20 terms is 820.
nth term is 4n-1
then
if n=1 ; 4(1)-1 = 3
n= 2 ; 4(2)-1 = 7
n= 3 ; 4(3)-1 = 11
----------------------------------
their common difference is 4
so that these are in the A.p.
3,7,11,...........................
the sum of nth term formula is
sn = n/2[2a+(n-1)d]
the sum of first 20 terms are
= 20/2[2(3)+(20-1)4]
= 10[6+(19)4]
= 10×[6+76]
= 10(82)
= 820.
the sum of the first 20 terms is 820.
srikrishnacharyulu:
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