In an AP, if a=1,an=20 and Sn= 399, then n=?
Devinitza:
Cn u make the question little clear?
Answers
Answered by
16
An=a+(n-1)d..............(1)
Sn=n/2[2a+(n-1)d]=n/2[a+An]..................(from (1))
399*2=n[1+20]
798/21=n
n=38
Sn=n/2[2a+(n-1)d]=n/2[a+An]..................(from (1))
399*2=n[1+20]
798/21=n
n=38
Answered by
11
WE HAVE- SN=399 , A=1 & AN=20
{Sn=n/2(a+an)}
=>so lets substitute...
399=n/2(a+an)
399=n/2(1+20)
399=n/2(21)
=>lets send 2 and 21 to RHS
399X2/21=n
=>then after the final calculation
38=n
i.e n=38
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HOPE THAT HELPS YOU
------------------PLZ RATE 5 STARS-------------
{Sn=n/2(a+an)}
=>so lets substitute...
399=n/2(a+an)
399=n/2(1+20)
399=n/2(21)
=>lets send 2 and 21 to RHS
399X2/21=n
=>then after the final calculation
38=n
i.e n=38
------
HOPE THAT HELPS YOU
------------------PLZ RATE 5 STARS-------------
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