In an AP, if Sn = n (4n + 1), find the AP.?
seenvasreddy:
plzz explain the first step clearly
why n-1 is written two times
u had explained well but still needs some explanation
respond to me
how this formula is substituted in to the sum
plzz explain step wise up to 8n-3 what u r thiking
Answers
Answered by
218
given ,
Sn =n ( 4n + 1 )
=4n^2 + n
we know ,
Tn = Sn - S(n-1)
=4n^2+n -4 (n-1)^2 - (n-1)
=4 (n^2-n^2+2n-1)+(n-n+1)
=8n - 4 + 1
= 8n -3
hence ,
Tn = 8n -3
T1 =8 (1)-3 =5
T2= 8 (2)-3 =13
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so, AP is 5, 13 , 21 ,...
Sn =n ( 4n + 1 )
=4n^2 + n
we know ,
Tn = Sn - S(n-1)
=4n^2+n -4 (n-1)^2 - (n-1)
=4 (n^2-n^2+2n-1)+(n-n+1)
=8n - 4 + 1
= 8n -3
hence ,
Tn = 8n -3
T1 =8 (1)-3 =5
T2= 8 (2)-3 =13
-----------
-------------
so, AP is 5, 13 , 21 ,...
dear where is two times
you can place (n-1) into n only
if 4n^2+n is given then place each term of n into (n-1)
now , 4 (n-1)^2 +(n-1)
this is easy
ya iam going thru this wait
yes that is fine in ur next step 4(n^2+n^2+2n-1) explan this
hi tell me now
taken 4 common there
(n-1)^2 that is( a-b)^2 is it ok
Answered by
157
Hope this helps you...
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