Question

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

• A. 96
• B. 78
• C. 91
• D. 84

Answer 1

Answer: (C)

91

Answer 2

Given:
$T_2 + T_6 = \frac{70}{3} \quad \text{and} \quad T_3 \cdot T_5 = 49.$ Let the first term of the geometric progression be $a$ and the common ratio be $r$.
The second term is:
$T_2 = ar,$ and the sixth term is:
$T_6 = ar^5.$

Given:
$ar + ar^5 = \frac{70}{3}.$ Factoring out $ar$:
$ar(1 + r^4) = \frac{70}{3}.$ (1)

The third term is:
$T_3 = ar^2,$and the fifth term is:
$T_5 = ar^4.$

Given:
$T_3 \times T_5 = ar^2 \times ar^4 = (ar^3)^2 = 49.$

Taking the square root:
$ar^3 = 7 \implies a = \frac{7}{r^3}.$ (2)

Substituting the value of $a$ from equation (2) into equation (1):
$\frac{7}{r^3} \times r \times (1 + r^4) = \frac{70}{3}.$

Simplifying:
$\frac{7}{r^2} (1 + r^4) = \frac{70}{3}.$

Multiplying both sides by $3r^2$:
$21(1 + r^4) = 70r^2.$

Rearranging terms:
$21 + 21r^4 = 70r^2.$

Dividing by 7:
$3 + 3r^4 = 10r^2.$

Letting $t = r^2$, we have:
$3 + 3t^2 = 10t.$

Rearranging:
$3t^2 - 10t + 3 = 0.$

Solving this quadratic equation using the quadratic formula:
$t = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6}.$ This gives:
$t = \frac{18}{6} = 3 \quad \text{or} \quad t = \frac{2}{6} = \frac{1}{3}.$

Since the GP is increasing, we take $t = 3$,

so:
$r^2 = 3 \implies r = \sqrt{3}.$ Using $r = \sqrt{3}$ in equation (2):
$a = \frac{7}{(\sqrt{3})^3} = \frac{7}{3\sqrt{3}} = \frac{7\sqrt{3}}{9}.$ Now, we find the sum of the 4th, 6th, and 8th terms:
$T_4 = ar^3, \quad T_6 = ar^5, \quad T_8 = ar^7.$ Calculating:
$T_4 + T_6 + T_8 = ar^3 + ar^5 + ar^7 = ar^3 (1 + r^2 + r^4).$ Substituting values:
$ar^3 = 7, \quad 1 + r^2 + r^4 = 1 + 3 + 9 = 13.$ Thus:
$T_4 + T_6 + T_8 = 7 \times 13 = 91.$ Therefore:
$91.$