In an AP, if a=1,an=20 and Sn= 399, then n=?
Answers
Answered by
603
since ,
an = a+(n-1)d
20 = 1 + (n-1)d
19 = (n-1)d
also ,
Sn = n/2 {2a+19}
399 = n/2(21)
n = 38
an = a+(n-1)d
20 = 1 + (n-1)d
19 = (n-1)d
also ,
Sn = n/2 {2a+19}
399 = n/2(21)
n = 38
Answered by
373
given,
a=1
an=20
sn=399
we know that,
an=a+(n-1)d
20=1+(n-1)d
20-1=(n-1)d
19= (n-1)d
sn=n/2{2n+(n-1)d}
399=n/2{2(1) + (n-1)d}
798=n(2+ 19)
798=n(21)
798/21=n
38=n
a=1
an=20
sn=399
we know that,
an=a+(n-1)d
20=1+(n-1)d
20-1=(n-1)d
19= (n-1)d
sn=n/2{2n+(n-1)d}
399=n/2{2(1) + (n-1)d}
798=n(2+ 19)
798=n(21)
798/21=n
38=n
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