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G.P. found from given products and sums of its terms

This type of question, asked 10 times: 2021 Paper 25, 26 Jun 2022, 28 Jun 2022, 29 Jan 2023, 13 Apr 2023, 8 Apr 2024, 22 Jan 2025, 3 Apr 2025, 7 Apr 2025, 21 Jan 2026 S1. All 10 questions below, oldest first.

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  1. Question 1

    2021 Paper 25

    66
    In an increasing, geometric series, the sum of the second and the sixth term is 252\frac{25}{2} and the product of the third and fifth term is 25. Then, the sum of 4th ,6th 4^{\text{th~}},6^{\text{th~}} and 8th 8^{\text{th~}} terms is equal to:

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  2. Question 2

    26 Jun 2022

    176
    If a1(>0),a2,a3,a4,a5a_{1}( > 0),a_{2},a_{3},a_{4},a_{5} are in a G.P., a2+a4=2a3+1a_{2}+a_{4}= 2a_{3}+ 1 and 3a2+a3=2a43a_{2}+a_{3}= 2a_{4}, then a2+a4+2a5a_{2}+a_{4}+ 2a_{5} is equal to

  3. Question 3

    28 Jun 2022

    66
    Let A1, A2, A3A_{1},{\text{ }A}_{2},{\text{ }A}_{3}, be an increasing geometric progression of positive real numbers. If A1A3A5A7=11296A_{1}A_{3}A_{5}A_{7}=\frac{1}{1296} and A2+A4=736A_{2}+A_{4}=\frac{7}{36}, then, the value of A6+A8+A10A_{6}+A_{8}+A_{10} is equal to

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  4. Question 4

    29 Jan 2023

    85
    Let a1,a2,a3,…a_{1},a_{2},a_{3},\ldots be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24 , then a1a9+a2a4a9+a5+a7a_{1}a_{9}+a_{2}a_{4}a_{9}+a_{5}+a_{7} is equal to

  5. Question 5

    13 Apr 2023

    155
    Let a1,a2,a3,…a_{1},a_{2},a_{3},\ldots be a G.P. of increasing positive numbers. Let the sum of its 6th 6^{\text{th~}} and 8th 8^{\text{th~}} terms be 2 and the product of its 3rd 3^{\text{rd~}} and 5th 5^{\text{th~}} terms be 19\frac{1}{9}. Then 6(a2+6\left( a_{2}+ \right.  a4)(a4+a6)\left. \ a_{4} \right)\left( a_{4}+a_{6} \right) is equal to

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  6. Question 6

    8 Apr 2024

    154
    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: -

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  7. Question 7

    22 Jan 2025

    57
    Let a1,a2,a3…a_{1},a_{2},a_{3}\ldots. be a G.P. of increasing positive terms. If a1a5=28a_{1}a_{5}= 28 and a2+a4=29a_{2}+a_{4}= 29, the a6a_{6} is equal to

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  8. Question 8

    3 Apr 2025

    66
    Let a1,a2,a3,…a_{1},a_{2},a_{3},\ldots be a G. P. of increasing positive numbers. If a3a5=729a_{3}a_{5}= 729 and a2+a4=1114a_{2}+a_{4}=\frac{111}{4}, then 24(a1+a2+a3)24\left( a_{1}+a_{2}+a_{3} \right) is equal to.

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  9. Question 9

    7 Apr 2025

    145
    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:

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  10. Question 10

    21 Jan 2026 S1

    56
    Let a1,a2,a3,…a_{1},a_{2},a_{3},\ldots. be a G.P. of increasing positive terms such that a2⋅a3⋅a4=64a_{2}\cdot a_{3}\cdot a_{4}= 64 and a1+a3+a5=8137a_{1}+a_{3}+a_{5}=\frac{813}{7}. Then a3+a5+a7a_{3}+a_{5}+a_{7} is equal to :

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