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The sum of first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is 13. Calculate the first and the thirteenth term.

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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.

Sum
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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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Chapter 9: Arithmetic and Geometric Progressions - Exercise 9.3

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ML Aggarwal Understanding Mathematics [English] Class 10 ICSE
Chapter 9 Arithmetic and Geometric Progressions
Exercise 9.3 | Q 15

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