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.

The sum of first 6 terms of an arithmetic progression is 42. The ratio

1.	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» 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.