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Question
The sum of the first 7 terms of an AP is 182. If its 4th and 17th terms are in the ratio 1:5, find the AP.
Solution
Let a be the first term and d be the common difference of the AP.
∴ s7 = 182
`⇒ 7/2 (2a + 6d) = 182 ( s_n = n/2 [ 2a + (n-1 ) d ]}`
⇒ a +3d = 26 ............. (1)
Also ,
a4 : a17 = 1:5 (Given)
`⇒ (a+3d)/(a+16d) = 1/5 [ a_n = a+(n-1) d]`
⇒ 5a +15d = a +16d
⇒ d = 4a .................(2)
Solving (1) and (2), we get
a+ 3× 4a = 26
⇒ 13a = 26
⇒ a = 2
Putting a =2 in (2), we get
d= 4 × 2 = 8
Hence, the required AP is 2, 10, 18, 26,………
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