JEE Main 2025 Apr 7 Shift 2, Mathematics Q20: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Apr 7, Shift 2Mathematics

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

A
760
B
755
C
750
D
757

Solution

Let the G.P. have first term $a$ and common ratio $r > 0$ . Then $a r + a r^{3} + a r^{5} = a r (1 + r^{2} + r^{4}) = 21$ , and $a r^{7} + a r^{9} + a r^{11} = a r^{7} (1 + r^{2} + r^{4}) = 15309$ .

Dividing the second equation by the first: $r^{6} = \frac{15309}{21} = 729 = 3^{6}$ , so $r = 3$ (since terms are positive).

Substituting back: $a (3) (1 + 9 + 81) = 21 \Rightarrow 273 a = 21 \Rightarrow a = \frac{1}{13}$ .

Sum of first 9 terms: $S_{9} = \frac{a ( r ^{9} - 1 )}{r - 1} = \frac{\frac{1}{13} ( 3 ^{9} - 1 )}{2} = \frac{19682}{26} = \frac{9841}{13} = 757$ .

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JEE Main 2025 Apr 7 Shift 2, Mathematics Q20: Arithmetic, Geometric And Harmonic Progressions

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