Can 240 of sum of a consecutive term is arithematic sequence of 7,11,15
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In the above AP, first term ‘a’= 7. Common difference ‘d’ = 4, Suppose total ’n' terms make the sum = 250
& Sn = n/2 * [ 2a + ( n-1)*d ] =
=> 250 = n/2 [ 14 + (n-1)*4 ]
=> 500 = n[ 14+4n-4] = n[10+4n]
=> 4n² + 10n - 500 = 0
=> 2n² + 5n - 250 = 0
=> 2n² +25n-20n -250 = 0
=> n(2n+25) - 10 (2n+ 25) = 0
=> ( 2n+ 25) (n-10 ) = 0
=> n = 10 & n= -25/2 ( which is ruled out, as n is no of terms)
=> n = 10
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