JEE Main 2022 · 28 Jun · Shift 1 · Mathematics Q6

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MathematicsSingle correct+4 / -1Question 6 of 90

Let $A_1,A_2,A_3,\dots$ be an increasing geometric progression of positive real numbers. If $A_1A_3A_5A_7=\dfrac1{1296}$ and $A_2+A_4=\dfrac7{36}$ , then, the value of $A_6+A_8+A_{10}$ is equal to

Answer and worked solution

Official answer: C

Product. $a^4r^{12}=\frac1{1296}$ , so $ar^3=\frac16$ . Then $ar=\frac7{36}-\frac16=\frac1{36}$ and $r^2=6$ .

Sum. $ar^5+ar^7+ar^9=\frac1{36}(36+216+1296)=43$ .

(C)

Topics:Geometric progression

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JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 6