Verified Solution • 2025–26 Session moderate

QJEEM19Mab321: JEE Main 2019 Mathematics - Class 11 JEEM19Mab321 Solution: Let , b and c be in G.P. with common ratio r, where 0 and 0 < r …

Let aa , b and c be in G.P. with common ratio r, where aa ≠\ne 0 and 0 < r ≤\le 12{1 \over 2} . If 3 aa , 7b and 15c are the first three terms of an A.P., then the 4th term of this A.P. is :

Multiple Choice Options

  • A aa
  • B 73a{7 \over 3}a
  • C 5 aa
  • D 23a{2 \over 3}a

Step-by-Step Verified Solution

a = a, b = ar and c = ar2

3a, 7b, 15c →\to A.P.

14b = 3a + 15c

14(ar) = 3a + 15(ar2)

15r2 – 14r + 3 = 0

⇒r=13,35(rejected)\Rightarrow r = {1 \over 3},{3 \over 5}(rejected)

Common difference = 7b – 3a

= 7ar – 3a

⇒\Rightarrow 7a3−3a=−23a{{7a} \over 3} - 3a = - {2 \over 3}a

4th term is ⇒\Rightarrow 15c−23a=159a−23a=a15c - {2 \over 3}a = {{15} \over 9}a - {2 \over 3}a = a

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