Verified Solution • 2025–26 Session moderate

QJEEM19Ma30b7: JEE Main 2019 Mathematics - Class 11 JEEM19Ma30b7 Solution: The product of three consecutive terms of a G.P. is 512. If 4…

The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is :

Multiple Choice Options

  • A 36
  • B 28
  • C 32
  • D 24

Step-by-Step Verified Solution

Let terms are ar,a,ar→G.P{a \over r},a,ar \to G.P

∴\therefore    a3{a^3} = 512   ⇒\Rightarrow  a = 8

8r+4,12,8r→A.P.{8 \over r} + 4,12,8r \to A.P.

24 = 8r+4+8r{8 \over r} + 4 + 8r

r = 2, r = 12{1 \over 2}

r = 2(4, 8, 16)

r = 12{1 \over 2} (16, 8, 4)

Sum = 28

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