IS this 3?
@Vocaloid
hm I got something different try writing out your an = a1 + (n-1)d formulas for the 7th term and the 10th term
the 7 is substituted for the d right
7 would be n
since we don't know a1 and d yet, leave those alone and substitute everything else
you should get one formula for a7 and one for a10
okay a(7)= a1 + (7-1)d a(10)=a1+(10-1)d
good, now substitute 21 for a7 and 126 for a10 since these are given
21 = a1 + (7-1)d 126 = a1 + (10-1)d now subtract the two equations. you should be able to eliminate a1.
each separate equations right
21=+69(d)
first subtract the left hand sides 126 - 21 then subtract the right hand sides a1 + (10-1)d - [a1 + (7-1)d]
let me know what you get for each side
3d
and 105
good and if 3d = 105, then d = ?
35
good, now we take that d value and substitute back into a7
21 = a1 + (7-1)(35) a1 = ?
189
should be negative -189 ^^ but yeah that's your answer
haha thank you
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