The sum of first 6 terms of an arithmetic progression is 42 the 10th term and 30th term are in the ratio 1 : 2 .Find the 13th term

1 thought on “The sum of first 6 terms of an arithmetic progression is 42 the 10th term and 30th term are in the ratio 1 : 2 .Find the 13th term”

1. Step-by-step explanation:

   Given,

   The sum of first 6 terms of an A.P = 42

   Ration of 10term of an A.P and 30th term of an A.P = 1 : 2

   To Find :–

   13th term of an A.P

   Formulas Required :–

   1) [tex]\sf a_n = a + (n – 1) d[/tex]

   2) [tex]\sf S_n = \dfrac{n}{2}[2a + (n – 1)d][/tex]

   Solution :–

   According to Question :-

   1) 42 = [tex]\sf\dfrac{6}{2}[2a + (6-1)d][/tex]

   42 = 3[2a + 5d]

   42 = 6a + 15d

   [Let it be Equation – 1]

   2) [tex]\sf\dfrac{a + (10-1)d}{a + (30 – 1)d} = \dfrac{1}{2}[/tex]

   [tex]\sf\dfrac{a + 9d}{a + 29d} = \dfrac{1}{2}[/tex]

   2[a + 9d] = 1[a + 29d]

   2a + 18d = a + 29d

   a = 11d

   Substituting value of ‘a’ in equation ‘1’ :-

   42 = 6(11d) + 15d

   42 = 66d + 15d

   42 = 81d

   d = 42/81 = 1/2

   a = 11d = 11(1/2) = 11/2

   13 th term of an A.P :-

   = [tex]\sf\left[\dfrac{11}{2}+(13-1)\dfrac{1}{2}\right][/tex]

   = [tex]\sf\left[\dfrac{11}{2}+6\right][/tex]

   = [tex]\sf\dfrac{23}{2}[/tex]

   [tex]\sf a_{13} = \dfrac{23}{2}[/tex]

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