the sum to infinity of a G.P series is R. The sum to infinity of the squares of the terms is 2R. The sum to infinity of the cubes of the terms is (64/13)R. Find (i) the value of R. (ii) the first term of the first original series.

Flambergru

Flambergru

Answered question

2022-08-02

The sum to infinity of a G.P series is R.
The sum to infinity of the squares of the terms is 2R.
The sum to infinity of the cubes of the terms is (64/13)R.
Find (i) the value of R. (ii) the first term of the first original series.

Answer & Explanation

choltas5j

choltas5j

Beginner2022-08-03Added 13 answers

When none of the terms are changed, let the first term be a and thecommon ratio be r.
When the terms are squared, the first term becomes a 2 and the common ratio becomes r 2
So,
a 1 r = R
Since
a 2 1 r 2 = 2 R
Then,
a 2 1 + r 2 = 2 a 1 r a 2 ( 1 + r ) ( 1 r ) = 2 a 1 r
Removing common terms,
a 1 + r = 2
a=2r+2
If the terms are cubed, the sum is
a 3 1 r 3 = 64 13 R
Substituting , a 3 1 r 3 = R
a 2 1 r 3 = 64 13 ( a 1 r ) a ( a 2 ) ( 1 r ) ( 1 + r + r 2 ) = 64 13 ( a 1 r )
Removing common terms,
a 2 r 2 + r + 1 = 64 13
Substituting a=2r+2=2(r+1)
8 ( r 2 + 2 r + 1 ) r 2 + r + 1 = 64 13
Dividing both sides by 8,
r 2 + 2 r + 1 r 2 + r + 1 = 8 13 Copyright ©2011-2012 CUI WEI. All Rights Reserved.
Cross-multiplying,
13 r 2 + 26 r + 14 = 8 2 + 8 r + 8 5 r 2 + 18 r + 5 = 0
After using the quadratic formula, you'll get r=-3.2967 or-0.3033
Since this is a infinite progression values of r with absolutevalues greater than one are not allowed. So, r cannot be-3.2967
Given r=-0.3033, a=2(-0.3033)+2>>a=1.393
So,
R=1.393/(1-(-0.3033))>>R=1.609

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