The sum of the first n terms of a geometric sequence is 255, and the

Question

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The sum of the first n terms of a geometric sequence is 255, and the sum of the reciprocals of these  n  terms is  \frac{255}{128} . The ratio,  r , of the sequence is a positive integer greater than 1, and the first term of the sequence is 1. The sum  S_{n} , of the first n terms of a geometric sequence is given by  S_{n}  =  \frac{a_{1}  }{1-r}  where  a_{1}  is the first term of the sequence.

In the table, select the number that equals the ratio, r, of the sequence, and the number that equals the number, n, of the terms. Make only one selection in each column.

bv8562 · Answer

Sum of first n terms = Sn = a1(1-r^n)/(1-r)

a1(1-r^n)/(1-r) = 255
(1-r^n)/(1-r) = 255

When we put r=2 and n=8 it satisfies the equation. Therefore, r=2 and n=8.

Aryaa03 · Answer

Sum of first n terms of geometric sequence =  S_{n}  =  \frac{a_{1}  }{1-r}   = 255 , and a1 = 1, r is the common ratio and n is the number of terms;
Sequence of the recriporcal of these terms will have a ratio of 1/r an therefore,
Sum of recriprocal of these terms =  S_{1/n}  = 1* (1- \frac{1}{r^n} )/(1- \frac{1}{r} )=  \frac{255}{128} 
Rearranging numerator and denominator,

S_{1/n}  =  \frac{1* }{1-r} *  \frac{1}{r^{n-1}}  =  \frac{255}{128}

Subsitute value of  \frac{1* }{1-r}  as 255 from first expression
==> 255* \frac{1}{r^{n-1}}  =  \frac{255}{128} 
 ==>  r^{n-1}  =  2^7 
=> r = 2 and n=8

Gemmie · Answer

(1) The sum of the first n terms of a geometric sequence is 255 
 a_1 * \frac{1 - r^n}{1-r}  =  \frac{1 - r^n}{1-r} = 255

(2) The sum of the reciprocals of these n terms is  \frac{255}{128}  
 \frac{1}{a_1} * (1 - (\frac{1}{r})^n) \div (1 - \frac{1}{r})

= (1 - 1\frac{}{r^n}) \div \frac{r-1}{r}

= \frac{r^n-1}{r^n} * \frac{r}{r-1}

= \frac{ }{r^{n-1}}  (because r^n - 1 = (r - 1) *  )

= \frac{255}{128}

128 = 2^7
=> n - 1 = 8
=> n = 8

Try r = 2
 1 * \frac{(1 - 2^8)}{(1-2)} = \frac{1 - 256}{-1} = 255 
=> r =2 is correct

VinitJain04 · Answer

How is the Rearranging numerator and denominator in the above solution is done? Can somebody simplify?

Sajjad1994 · Answer

Official Explanation The first n terms of the sequence are 1, r, r^2, r^3, . . . , r^n . The reciprocals of these terms are 1, \frac{1}{r}. \frac{1}{r^2}. \frac{1}{r^3}. ......\frac{1}{r^n} , These terms form a geometric sequence, as well, with ratio Now, apply the formula for the sum of terms of a geometric sequence twice, for the original sequence and for the sequence of the inverses. First, the sum of the terms of the original sequence: https://gmatclub.com/forum/download/file.php?id=82013 Clearly, if r equals 2 and n – 1 equals 7 (that is, n equals 8), the above expression is correct. These two values are in the answer choices, so they have to be the right answers. GMAT-Club-Forum-6a9w8457.jpeg

Yosemite98 · Answer

We know that the sequence will be:

a1 = 1
a2 = r
a3 = r^2
a4 = r^3
..
an = r^(n-1)

the sum of the reciprocals will be = 1+ (1/r)+(1/r^2) +(1/r^3)... +(1/r^(n-1)) = r^(n-1)+ r^(n-2)+r^(n-3)+....1/(r^(n-1)) = 255/128

We can see that the numerator is the sum of the sequence = 255 so that means that the denominator should be equal to 128 = r^(n-1) = 2^7-> this is satisfied by r= 2, n=8

vibhorrokstar19 · Answer

I used a different approach here ! As it is given in the question we can form the equation for the sum of the integers in GP. Then after further simplifying we get the equation :- 254 = 255r - r^n 
we can then factorise 254 which is 2*127 
therefore 2*127=r(255-r^n-1) since 127 is prime so 255-r^n-1 can be 127 when r^n-1 is 128. we have the answer as N=8 and R=2.

The sum of the first n terms of a geometric sequence is 255, and the

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