Page 236 - aljaber

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.n = 18 .1 1

Sn = n  2a1 + (n − 1)d 
2

171 = n  2 × 1 + (n − 1)× 1
2
342 = n(1 + n)

342 = n + n2

n2 + n − 342 = 0

(n + 19)(n − 18) = 0 n = −19

. ()
. 1 n(n + 1)
n
2

a1 = 1 1, 2, 3, …, n

.Sn = 1 + 2 + 3 + .d = 1 an = n

17 21 + n = n (1 + n) = n (n + 1)
22

()

. .0

a6 = a1 + 5d ⇒ 21 = a1 + 5d

S17 = 17 (2a1 + 16d) ⇒ 0 = 17 (2a1 + 16d)
2 2