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EXERCISE 2.3 · Q57

Q.The sum of an infinite G.P. is 5 and the sum of the squares of these terms is 15. Find the G.P.

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a1−r=5⇒a=5(1−r)\dfrac a{1-r}=5 \Rightarrow a=5(1-r). Sum of squares: a21−r2=15\dfrac{a^2}{1-r^2}=15. Substituting: 25(1−r)2(1−r)(1+r)=15⇒25(1−r)1+r=15⇒25−25r=15+15r⇒10=40r⇒r=14\dfrac{25(1-r)^2}{(1-r)(1+r)}=15 \Rightarrow \dfrac{25(1-r)}{1+r}=15 \Rightarrow 25-25r=15+15r \Rightarrow 10=40r \Rightarrow r=\dfrac14. Then $a=5\le …

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