How many terms of the AP 17, 15, 13, 11, ... must be added to get the sum 72? Explain the double answer.

Explanation

Here,a = 17 and d = (15-17) =-2.
Let the sum of n terms be 72. Then,
`S_(n) = 72 rArr (n)/(2) * {2a + (n-1)d} = 72`
`rArr n * {2 xx 17 + (n-1) xx (-2)} = 144`
`rArr n(36-2n) = 144 rArr 2n^(2) -36n + 144 =0`
`rArr n^(2) -18n +72 = 0 rArr n^(2) -12n -6n +72 =0`
`rArr n(n-12) -6(n-12) = 0 rArr (n-12)(n-6) = 0`
`rArr n = 6 "or" n = 12.`
`therefore ` sum of first 6 terms = sum of first 12 terms = 72.
This means that the sum of all terms from 7th to 12th is zero.