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Question
The sum of 15 terms of an A.P. is zero. Its 4th term is 12. Find its 14th term.
Sum
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Solution
`S_n = n/2[2a + (n - 1)d]`
`S_15 = 15/2[2a + (15 - 1)d]`
2a + 14d = 0
2a = −14d
a = `(−14d)/2`
a = −7d
an = a + (n − 1)d
an = a + (4 − 1)d
a + 3d = 12
−7d + 3d = 12
−4d = 12
d = `12/-4`
d = −3
a = −7d
= (−7) × (−3)
= 21
a14 = a + (14 − 1)d
= 21 + 13 × (−3)
= 21 − 39
= −18
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