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Question
If sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is –3 then find the 10th term.
Solution
Let the first term and common difference of an AP are a and d, respectively.
According to the questiuon,
a3 + a8 = 7 and a7 + a14 = –3
⇒ a + (3 – 1)d + a + (8 – 1)d = 7 ...[∵ an = a + (n – 1)d]
And a + (7 – 1)d + a + (14 – 1)d = –3
⇒ a + 2d + a + 7d = 7
And a + 6d + a + 13d = –3
⇒ 2a + 9d = 7 ...(i)
And 2a + 19d = –3 ...(ii)
On subtracting equation (i) from equation (ii), we get
10d = –10
⇒ d = –1
2a + 9(–1) = 7 ...[From equation (i)]
⇒ 2a – 9 = 7
⇒ 2a = 16
⇒ a = 8
∴ a10 = a + (10 – 1)d
= 8 + 9(–1)
= 8 – 9
= –1
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