Question

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 of an arithmetic progression​

Answer 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) $\sf a_n = a + (n - 1) d$

2) $\sf S_n = \dfrac{n}{2}[2a + (n - 1)d]$

Solution :-

According to Question :-

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

42 = 3[2a + 5d]

42 = 6a + 15d

[Let it be Equation - 1]

2) $\sf\dfrac{a + (10-1)d}{a + (30 - 1)d} = \dfrac{1}{2}$

$\sf\dfrac{a + 9d}{a + 29d} = \dfrac{1}{2}$

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 :-

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

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

= $\sf\dfrac{23}{2}$

$\sf a_{13} = \dfrac{23}{2}$