Question

The sum of first 6 terms of an arithmetic progression is 42. The ratio of its 10th term to its 30th term is `1:3`. Calculate the first and 13th term of an AP.

Answer 1

Let a be the first term and d the common difference of the given AP. Then,
`T_(10) = a + 9d "and" T_(30) = a +29d`
`therefore (T_(10))/(T_(30)) = (1)/(3) rArr (a+9d)/(a +29d) = (1)/(3)`
`rArr 3a + 27d = a +29d`
`rArr 2a = 2d rArr a =d.`
`"Also,"S_(n) = (n)/(2) [2a + (n-1)d]`
`rArr S_(6) = (6)/(2)(2a +5d)`
` = (6a + 15d) = (6a + 15a) " " [because d =a]`
= 21a
But, `S_(6) = 42` (given)
`therefore 21a = 42 rArr a = 2.`
Thus, a = 2 and d = 2.
`therefore "13th term," T_(13) = (a+12d) = (2 +12 xx 2) = 26`
Hence, the first term is 2 and the 13th term is 26.