Questions
The minimum number of terms from the beginning of the series so that the sum may
exceed 1568, is
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a
25
b
27
c
28
d
29
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Correct option is D
It is in A.P. for which a=20,d=223=83Now, Sn>1568⇒ n240+(n−1)83>1568⇒ n2×112+8n3>1568⇒ n2+14n>68×1568=1176⇒ n2+14n−1176>0 or (n+42)(n−28)>0As n is positive, n – 28 > 0 i.e., n > 28 ∴Minimum value of n = 29.Similar Questions
If the sum of first terms of an is , then the sum of squares of these terms is
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