CBSE 10th Class past year question paper
Year 2019
SET 1, Code Number- 30/2/1
Question number – 29 (b)
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Sn = n/2 × { 2a + ( n – 1 ) d }
180 = n/2 × [90 (n – 1)(–6)]
360 = 90n – 6n² + 6n
6n² – 96n + 360 = 0
6 ( n² – 16 + 60 ) = 0
n² – 16 + 60 = 0
n² – 10n – 6n + 60 = 0
n ( n – 10 ) – 6 (n – 10 ) = 0
( n – 6 ) ( n – 10 ) = 0
n = 6, n = 10
∴ n = 6 or n = 10