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JOHAN (CA-FINAL) (277 Points)

18 May 2012

Q1 If 5th term of a G.P. IS 3√3,then the product of first nine terms is

<a>8 <b>27 <c>243 <d>9.........

Q2 Sum of series 1+ 4/5+7/5²+10/5³+........infinity is

<a>15/36 <b>35/36 <c>35/16 <d>15/16........

Also show explanation to both ans..........

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2 Replies

Tejaswi Kasturi (student-cpt) (427 Points)
Replied 18 May 2012

Let the terms of the G.P. be a/r⁴,a/r³,a/r²,a/r,a,ar,ar²,ar³,ar⁴ These are the 9 terms in G.P. fifth term is a , a= 3√3 product of nine terms is a^9 = 1594323√3 There is no option in the answer
The n^th term is (3*n – 2)/5ⁿ

∑(3*n – 2)/5ⁿ = 3*^inf∑_n=1(n/5ⁿ) -2*^inf∑_n=1(1/5ⁿ)

The sum of the second sigma is 5/4 (infinite G.P. 1/(1-1/5) ) let first sigma be denoted by S

S = ^inf∑_n=1(1/5ⁿ) + ^inf∑_n=1(n-1/5ⁿ) ( since n = (n-1) + 1 )

The first sigma in S is 5/4. Let the second sigma be S₁. Now S₁ = ^inf∑_n=1(n-1/5ⁿ) but notice that the first term of S₁ is 0 as n-1=0 if n=1 so S₁ can be written as ^inf∑_n=2(n-1/5ⁿ) now instead of n substitute t = n-1 in the equation of S₁. Now

S₁ = ^inf∑_t=1(t/5^t+1)

Taking 1/5 as common

S₁ =1/5( ^inf∑_t=1(t/5^t)) = S/5 (notice that the expression in brackets is same as the original S)

so S = 5/4 + S/5 implies 4S/5 = 5/4 implies S = 25/16

substituting in the first equation we get

sum = 3*25/16 – 2*5/4 = 35/16

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Tejaswi Kasturi (student-cpt) (427 Points)
Replied 19 May 2012

In the above answer for the second question given by me the answer was correct but the method with which I solved the sum is wrong. My first mistake was that the term t_n is (3*n-2)/5^n-1. Now to sum it…
∑(3*n – 2)/5^n-1 =15*^inf∑_n=1(n/5ⁿ) -10*^inf∑_n=1(1/5ⁿ)

My second mistake was to say that the sum of second sigma is 5/4. It is not 5/4 but it is 1/4 .

S₀ = 1/5(1/(1 – 1/5)) = ¼ (I forgot that the initial term is 1/5 and not 1)

let first sigma be denoted by S

S= ^inf∑_n=1(1/5ⁿ)+ ^inf∑_n=1(n-1/5ⁿ)(since n = (n-1)+1)

The first sigma is S is ¼ Let the second sigma be S₁. Now S₁ = ^inf∑_n=1(n-1/5ⁿ) but notice that the first term of S₁ is 0 as n-1=0 if n=1 so S₁ can be written as ^inf∑_n=2(n-1/5ⁿ) now instead of n substitute t = n-1 in the equation of S₁. Now