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Question
The sum of first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is `(1)/(3)`. Calculate the first and the thirteenth term.
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Solution
T10 : T30 = 1 : 3, S6 = 42
Let a be the first term and d be a common difference, then
`(a + 9d)/(a + 29d) = (1)/(3)`
⇒ 3a + 27d = a + 29d
⇒ 3a – a = 29d – 27d
⇒ 2a = 2d
⇒ a = d
Now, S6 = 42
= `n/(2)[2a + (n - 1)d]`
⇒ 42 = `(6)/(2)[2a + (6 - 1)d]`
⇒ 42 = 3[2a + 5d]
⇒ 14 = 2a + 5d
⇒ 14 = 2a + 5a ...(∵ d = a)
⇒ 7a = 14
⇒ a = `(14)/(7)` = 2
∴ a = d = 2
Now, T13 = a + (n – 1)d
= 2 + (13 – 1) x 2
= 2 + 12 x 2
= 2 + 24
= 26
∴ 1st term is 2 and thirteenth term is 26.
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