If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio

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In the given problem, we are given an A.P whose 18th and 11th term are in the ratio 3:2 We need to find the ratio of its 21st and 5th terms Now, using the formula an = a + ( n -1) d Where, a = first tem of the A.P n = number of terms d = common difference of the A.P So, a18 = a + ( 18 - 1) d a18 = a + 17d Also, a11 = a + ( 11 -1 ) d a11 = a + 10 d Thus, (a_18)/(a_11) = $\frac{3}{2}$ `( a + 17d) / ( a + 10 d) = $\frac{3}{2}$` 2 (a + 17 d ) = 3 ( a + 10d) 2a + 3ad = 3a + 30d Further solving for a, we get 34 d - 30d = 3a - 2a 4d = a .............(1) Now, a21 = a + (21 - 1) d a21 = a + 20 d Also, a5 = a +( 5 -1) d a5 = a + 4d So, `a_21/a_5 = ( a + 20d ) /( a + 4d) ` Using (1) in the above equation, we get `a_21/a_5 = ( 4d + 20 d) / (4d + 4d) a_21/a_5 = (24d)/(8d) a_21 / a_5 = $\frac{3}{1}$` Thus, the ratio of the 21st and 5th term is 3: 1 .

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