Mathematics

The product of n terms of a G.P. with first term a and common ratio r, r > 1 is :
$\dfrac{a^n(r^{n} - 1)}{r - 1}$
aⁿrⁿ
$a^{n} r^{\dfrac{n(n+1)}{2}}$
$a^{n} r^{\dfrac{n(n-1)}{2}}$

G.P.

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Answer

Given,

G.P. : a, ar, ar², ar³,…..,ar^{n - 1}

Product = a × ar × ar² × ar³×…..×ar^{n - 1}

= aⁿ × r^{0 + 1 + 2 + 3 + …. + (n-1)}

= $a^nr^\dfrac{n(n-1)}{2}$ [As sum of first n - 1 natural numbers = $\dfrac{n(n - 1)}{2}$ ]

Hence, option 4 is the correct option.

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Consider the G.P. a, ar, ar², …….., l. The kth term from the end is :
$\dfrac{l}{r^{k-1}}$
$\dfrac{l}{k}$
$\dfrac{al}{r^k}$
$\dfrac{l}{ar^{k-1}}$
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If a and l are respectively the first and the last terms of a G.P. having common ratio r > 1, then the sum to n terms of the G.P. is given by :
lar^{n - 1}
ar^ln
$\dfrac{lr - a}{r - 1}$
$\dfrac{lr^n - a}{lr - 1}$
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Case Study
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, then
Based on the above given information, answer the following questions:
The number of bacteria that would be present at the end of 4th hour is :
(a) 240
(b) 480
(c) 960
(d) 450
The number of bacteria that would be present at the end of 10th hour is :
(a) 30720
(b) 15360
(c) 61440
(d) 27540
The number of bacteria that would be present at the end of nth hour is :
(a) 30 × 2^{(n - 1)}
(b) 30 × 2^{(n + 1)}
(c) 30 × 2ⁿ
(d) none of these
What would be the sum of the first 5 terms of the G.P. that is formed by the number of bacteria after the completion of each hour?
(a) 450
(b) 930
(c) 1890
(d) none of these
What would be the sum of the first n terms of the G.P. that is formed by the number of bacteria after the completion of each hour?
(a) 30 × 2⁽ⁿ⁾ - 1
(b) 30 × (2^{(n - 1)} - 1)
(c) 30 × (2^{(n + 1)} - 1)
(d) 30 × 2ⁿ - 30
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Case Study
A man writes a letter to four of his friends. He asks each one of them to copy the letter and mail it to four different persons with the instruction that they move the chain similarly. Assume that the chain is not broken and it costs ₹4 to mail one letter.
Based on the above given information, answer the following questions:
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(a) 2048
(b) 8192
(c) 4096
(d) 1024
2.The amount spent on the postage of 6th set of letters is :
(a) ₹ 16,384
(b) ₹ 8,192
(c) ₹ 32,768
(d) ₹ 4,096
3.The total number of letters mailed till the 5th set of letters is :
(a) 1024
(b) 4092
(c) 1236
(d) 1364
4.The amount spent on the postage till the 5th set of letters is :
(a) ₹ 16,368
(b) ₹ 4,096
(c) ₹ 5,456
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(b) ₹ $\Big(\dfrac{16}{3}(4^n - 1)\Big)$

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Mathematics

The product of n terms of a G.P. with first term a and common ratio r, r > 1 is :an(rn−1)r−1\dfrac{a^n(r^{n} - 1)}{r - 1}r−1an(rn−1)​anrnanrn(n+1)2a^{n} r^{\dfrac{n(n+1)}{2}}anr2n(n+1)​anrn(n−1)2a^{n} r^{\dfrac{n(n-1)}{2}}anr2n(n−1)​

G.P.

Answer

Related Questions

Consider the G.P. a, ar, ar2, …….., l. The kth term from the end is :lrk−1\dfrac{l}{r^{k-1}}rk−1l​lk\dfrac{l}{k}kl​alrk\dfrac{al}{r^k}rkal​lark−1\dfrac{l}{ar^{k-1}}ark−1l​

If a and l are respectively the first and the last terms of a G.P. having common ratio r > 1, then the sum to n terms of the G.P. is given by :larn - 1arlnlr−ar−1\dfrac{lr - a}{r - 1}r−1lr−a​lrn−alr−1\dfrac{lr^n - a}{lr - 1}lr−1lrn−a​

The product of n terms of a G.P. with first term a and common ratio r, r > 1 is :
$\dfrac{a^n(r^{n} - 1)}{r - 1}$
aⁿrⁿ
$a^{n} r^{\dfrac{n(n+1)}{2}}$
$a^{n} r^{\dfrac{n(n-1)}{2}}$

Consider the G.P. a, ar, ar², …….., l. The kth term from the end is :
$\dfrac{l}{r^{k-1}}$
$\dfrac{l}{k}$
$\dfrac{al}{r^k}$
$\dfrac{l}{ar^{k-1}}$

If a and l are respectively the first and the last terms of a G.P. having common ratio r > 1, then the sum to n terms of the G.P. is given by :
lar^{n - 1}
ar^ln
$\dfrac{lr - a}{r - 1}$
$\dfrac{lr^n - a}{lr - 1}$