(vi) given a = 2, d=8, Sn= 90, find n and an .
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a = 2
d = 8
Sn = 90
Sn = n/2 [ 2a + ( n – 1 )d ]
90 = n/2 [ 2×2 + ( n – 1 ) 8 ]
90 = n/2 × 2 [ 2 + 4n – 4 ]
90 = n [ – 2 + 4n ]
45 = n [ – 1 + 2n ]
45 = – n + 2n²
2n² – n – 45 = 0
2n² –10n + 9n – 45 = 0
2n ( n – 5 ) + 9 ( n – 5 ) = 0
( n – 5 ) ( 2n + 9 ) = 0
n = 5 [ n ≠ –9/2 since no of terms cannot be negative ]
an = a + ( n – 1 ) d
an = 2 + 4d
an = 2 + 4 × 8
an = 2 + 32
an = 34
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