Verified Solution • 2025–26 Session moderate

QJEEM24Ma4dc7: JEE Main 2024 Mathematics - Class 11 JEEM24Ma4dc7 Solution: In an increasing geometric progression of positive terms, the…

In an increasing geometric progression of positive terms, the sum of the second and sixth terms is 703\frac{70}{3} and the product of the third and fifth terms is 49. Then the sum of the 4th ,6th 4^{\text {th }}, 6^{\text {th }} and 8th 8^{\text {th }} terms is equal to:

Multiple Choice Options

  • A 78
  • B 96
  • C 91
  • D 84

Step-by-Step Verified Solution

Let's denote the first term of the geometric progression by aa and the common ratio by rr . The terms of the geometric progression can be written as follows:

First term: aa

Second term: arar

Third term: (ar2)(ar^2)

Fourth term: (ar3)(ar^3)

Fifth term: (ar4)(ar^4)

Sixth term: (ar5)(ar^5)

Eighth term: (ar7)(ar^7)

We are given two key pieces of information:

1. The sum of the second and sixth terms is 703\frac{70}{3} :

ar+ar5=703ar + ar^5 = \frac{70}{3}

2. The product of the third and fifth terms is 49:

(ar2)⋅(ar4)=49(ar^2) \cdot (ar^4) = 49

a2r6=49a^2 r^6 = 49

a2=49r6a^2 = \frac{49}{r^6}

a=7r3a = \frac{7}{r^3}

Substituting a=7r3a = \frac{7}{r^3} into the first equation:

7r3⋅r+7r3⋅r5=703\frac{7}{r^3} \cdot r + \frac{7}{r^3} \cdot r^5 = \frac{70}{3}

7rr3+7r5r3=703\frac{7r}{r^3} + \frac{7r^5}{r^3} = \frac{70}{3}

7r2+7r21=703\frac{7}{r^2} + \frac{7r^2}{1} = \frac{70}{3}

Let x=r2x = r^2 . Then:

7x+7x=703\frac{7}{x} + 7x = \frac{70}{3}

Multiply through by 3x to clear the denominator:

21+21x2=70x21 + 21x^2 = 70x

Rearrange into a standard quadratic equation:

21x2−70x+21=021x^2 - 70x + 21 = 0

Divide by 7 to simplify:

3x2−10x+3=03x^2 - 10x + 3 = 0

Solve this quadratic equation using the quadratic formula:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Where a=3a = 3 , b=−10b = -10 , and c=3c = 3 . Thus:

x=10±100−366x = \frac{10 \pm \sqrt{100 - 36}}{6}

x=10±646x = \frac{10 \pm \sqrt{64}}{6}

x=10±86x = \frac{10 \pm 8}{6}

x=3x = 3 or x=13x = \frac{1}{3}

Since x=r2x = r^2 and rr is positive, we get r=3r = \sqrt{3} or r=13r = \frac{1}{\sqrt{3}} . We need to choose the value that results in positive, increasing terms:

If r=3r = \sqrt{3} :

a=7r3=7(3)3=733=73⋅13=739a = \frac{7}{r^3} = \frac{7}{(\sqrt{3})^3} = \frac{7}{3\sqrt{3}} = \frac{7}{3} \cdot \frac{1}{\sqrt{3}} = \frac{7\sqrt{3}}{9}

Now we can determine the sum of the 4th, 6th, and 8th terms:

The 4th term is: ar3=739⋅33=7ar^3 = \frac{7\sqrt{3}}{9} \cdot 3\sqrt{3} = 7

The 6th term is: ar5=739⋅93=21ar^5 = \frac{7\sqrt{3}}{9} \cdot 9\sqrt{3} = 21

The 8th term is: ar7=739⋅27(3)=49ar^7 = \frac{7\sqrt{3}}{9} \cdot 27(\sqrt{3}) = 49

Adding these together:

(4th+6th+8thterms)=7+21+63=91(4th + 6th + 8th terms) = 7 + 21 + 63 = 91

Therefore, the sum of the 4th, 6th, and 8th terms is 91.

Correct answer:

Option C: 91

Related Learning Resources & Exam Preparation