find the sum of the terms of an A. P. 2,6,0,114........,32
Anonymous:
hey friend, the AP is 2,6,10,14 ... 32
please repost your question with correct values
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AP = 2,6,10,14,......34
Here,
a= 2 , d = t2-t1 = 6-2= 4
an= 34.
Sn= ?
an= a+ ( n-1) × d
34= a + (n-1) × 4
34 = 2 + 4n -4
34-2+4= 4n
36= 4n
n= 36/4= 9
we know that,
Sn = n/2 ( 2a + ( n-1) × d
put all the value u will get ur answer
Here,
a= 2 , d = t2-t1 = 6-2= 4
an= 34.
Sn= ?
an= a+ ( n-1) × d
34= a + (n-1) × 4
34 = 2 + 4n -4
34-2+4= 4n
36= 4n
n= 36/4= 9
we know that,
Sn = n/2 ( 2a + ( n-1) × d
put all the value u will get ur answer
it is right
No dear... the AP is 2,6,10,14
in my book it is written like that only
so please can you help me
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