An infinite geometric series has sum 50. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is ( m / n ) where m, n are relatively prime integers. Find the value of ( m + n ).

Question

An infinite geometric series has sum 50. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is ( m / n ) where m, n are relatively prime integers. Find the value of ( m + n ). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let a, a r, a r2, ldots ldots . .. be an infinite G.P. So, ( a /1- r )=50....(1) Also, ( a 2/1- r 2)=10 ⋅ 50 ⇒ ( a /1+ r )=10 .....(2) [Using equation (1)] ∴ ( text Equation (1) / text Equation (2))=(1+r/1-r)=5 ⇒ 1+r=5-5 r ⇒ 6 r=4 ⇒ r=(2/3) ∴ m+n=5

Q. An infinite geometric series has sum 50. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is nm​ where m, n are relatively prime integers. Find the value of (m+n).

Let a,ar,ar2,……... be an infinite G.P. So, 1−ra​=50....(1) Also, 1−r2a2​=10⋅50⇒1+ra​=10 .....(2) [Using equation (1)] ∴ Equation (2) Equation (1) ​=1−r1+r​=5⇒1+r=5−5r⇒6r=4⇒r=32​ ∴m+n=5

Q. An infinite geometric series has sum 50. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $\frac{m}{n}$ where $m$ , $n$ are relatively prime integers. Find the value of $(m + n)$ .