Let A1, A2, A3, ....... be an increasing geometric progression of positive real numbers. If A1A3A5A7 = 1 \over 1296 and A2 + A4 = 7 \over 36 , then the value of A6 + A8 + A10 is equal to

Sequences and Series PYQ — JEE Main 2022 · 28 Jun · Shift 1 · Q27 with Solution

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Correct answer: C

Let the increasing geometric progression be

\quad A_2=ar, \quad A_3=ar^2,\dots$$ with $$a>0,\quad r>1$$ because the GP is increasing and all terms are positive. ## 1. Use the product condition Given $$A_1A_3A_5A_7=\frac1{1296}.$$ Now, $$A_1=a, \quad A_3=ar^2, \quad A_5=ar^4, \quad A_7=ar^6.$$ So $$A_1A_3A_5A_7=a\cdot ar^2\cdot ar^4\cdot ar^6=a^4r^{12}=(ar^3)^4.$$ Hence $$(ar^3)^4=\frac1{1296}.$$ Since $$1296=6^4,$$ we get $$ar^3=\frac16.$$ But note that $$A_4=ar^3,$$ so $$A_4=\frac16.$$ ## 2. Use the sum condition Given $$A_2+A_4=\frac7{36}.$$ Now, $$A_2=ar,\quad A_4=ar^3=\frac16.$$ Thus $$ar+\frac16=\frac7{36}.$$ So $$ar=\frac7{36}-\frac16=\frac7{36}-\frac6{36}=\frac1{36}.$$ Hence $$A_2=\frac1{36}.$$ ## 3. Find the common ratio We have $$A_4=A_2r^2.$$ So $$\frac16=\frac1{36}r^2.$$ Therefore $$r^2=6.$$ Since the GP is increasing, $r>1$, hence $$r=\sqrt6.$$ ## 4. Find $A_6+A_8+A_{10}$ Now, $$A_6=A_4r^2=\frac16\cdot 6=1,$$ $$A_8=A_6r^2=1\cdot 6=6,$$ $$A_{10}=A_8r^2=6\cdot 6=36.$$ Therefore, $$A_6+A_8+A_{10}=1+6+36=43.$$ ## 5. Check options The value is $$43,$$ which corresponds to **Option C**. ## 6. Comparison with stored answer Stored correct answer: **C** Our derived answer: **C** So the answer agrees with the stored correct answer.

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Correct answer: C

Let the increasing geometric progression be

\quad A_2=ar, \quad A_3=ar^2,\dots$$ with $$a>0,\quad r>1$$ because the GP is increasing and all terms are positive. ## 1. Use the product condition Given $$A_1A_3A_5A_7=\frac1{1296}.$$ Now, $$A_1=a, \quad A_3=ar^2, \quad A_5=ar^4, \quad A_7=ar^6.$$ So $$A_1A_3A_5A_7=a\cdot ar^2\cdot ar^4\cdot ar^6=a^4r^{12}=(ar^3)^4.$$ Hence $$(ar^3)^4=\frac1{1296}.$$ Since $$1296=6^4,$$ we get $$ar^3=\frac16.$$ But note that $$A_4=ar^3,$$ so $$A_4=\frac16.$$ ## 2. Use the sum condition Given $$A_2+A_4=\frac7{36}.$$ Now, $$A_2=ar,\quad A_4=ar^3=\frac16.$$ Thus $$ar+\frac16=\frac7{36}.$$ So $$ar=\frac7{36}-\frac16=\frac7{36}-\frac6{36}=\frac1{36}.$$ Hence $$A_2=\frac1{36}.$$ ## 3. Find the common ratio We have $$A_4=A_2r^2.$$ So $$\frac16=\frac1{36}r^2.$$ Therefore $$r^2=6.$$ Since the GP is increasing, $r>1$, hence $$r=\sqrt6.$$ ## 4. Find $A_6+A_8+A_{10}$ Now, $$A_6=A_4r^2=\frac16\cdot 6=1,$$ $$A_8=A_6r^2=1\cdot 6=6,$$ $$A_{10}=A_8r^2=6\cdot 6=36.$$ Therefore, $$A_6+A_8+A_{10}=1+6+36=43.$$ ## 5. Check options The value is $$43,$$ which corresponds to **Option C**. ## 6. Comparison with stored answer Stored correct answer: **C** Our derived answer: **C** So the answer agrees with the stored correct answer.