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Question
Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a4 = 9
⇒ a + (4-1) d = 9 [ an = a + (n-1) d]
⇒ a +3d = 9 ....................(1)
Now,
a6 +a13 = 40 (Given)
⇒ (a +5d ) + (a +12d) = 40
⇒ 2a + 17d = 40 ...............(2)
From (1) and (2), we get
2(9-3d ) +17d = 40
⇒ 18-6d + 17d = 40
⇒ 11d = 40 - 18 =22
⇒ d =2
Putting d = 2 in (1), we get
a +3 × 2 = 9
⇒ a = 9-6=3
Hence, the AP is 3, 5, 7, 9, 11,…….
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