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Question
The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.
Sum
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Solution
Let a and d the first term and common difference of an AP, respectively.
Now, by given condition,
a8 = `1/2 a_2`
⇒ a + 7d = `1/2 (a + d)` ...[∵ an = a + (n – 1)d]
⇒ 2a + 14d = a + d
⇒ a + 13d = 0 ...(i)
And a11 = `1/3 a_4 + 1` ...[Given]
⇒ a + 10d = `1/3[a + 3d] + 1`
⇒ 3a + 30d = a + 3d + 3
⇒ 2a + 27d = 3
From equations (i) and (ii),
2(–13d) + 27d = 3
⇒ –26d + 27d = 3
⇒ d = 3
From equation (i),
a + 13(3) = 0
⇒ a = – 39
∴ a15 = a + 14d
= – 39 + 14(3)
= – 39 + 42
= 3
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