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a1,a2,a3an are in A.P such that an=100,a50a49=3/5 then 15th  term of A.P from the end is

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a
4545
b
4485
c
4525
d
4585

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detailed solution

Correct option is D

Let T15 is 15th  term from end T15=an−14d=100-1435=100−425=4585


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The sum of n terms of an A.P. is 3n2+5 The number of  term which equals 159, is


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