Q2 IUustrat•on 5:
term or the sequence —6, —2, 2,
is 394?
Solution: First an observe that the dmerence Of utive
common. So is an Ap.
Its first tem a = —6
Common difference = 4
Therefore the general term is given by
4n—4
or a, = an — 10
Let an = 394
so, 4n - 10 • 394
Thus 4n = 394 10.404
Hence n 404/4 101.
i.e. 394 is 101st term.
:cercise
Smwthatme sequence 4n omA.P, Whet is itecommor,
Let an... bean AP. wth
AP wth Prove toat is
Find ts cumm afference.
Two AP_s have the sarne com-non difference. Fb-st term o • , one of these