Question

║⊕QUESTION⊕║
As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality.
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CLASS 11
SEQUENCES AND SERIES
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Find three numbers in GP whose sum is 35 and the sum of their squares is 525

Answer 1

Hii !!

Solution is in this given picture !

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@Rajukumar111

Answer 2

Answer: 20 , 10 , 5

Step-by-step explanation:

$\text{Let the numbers be :}\ a,ar,ar^2 \\ \\ \text{Sum of terms in GP =}\ 35 = a(1+r+r^2)\\ \\ \text{Squaring we get :}\\ \\ a^2(1+r+r^2)^2 = 35^2 = 1225 \implies \ (i) \\ \\ \text{Sum of square of terms :} \\ \\ a^2(1+r^2+r^4) = 525 = a^2(1+r+r^2)(1-r+r^2) \implies \ (ii) \\ \\ \text{Subtracting 2 from 1 we get:} \\ \\ a^2(1+r+r^2)-a^2(1+r+r^2)(1-r+r^2) = 1225 - 525 \\ \\ a^2(1+r+r^2)\big [(1+r+r^2)-(1-r+r^2)\big] = 700 \\ \\ 2(a^2r)(1+r+r^2)= 700 \implies 2(ar)\big(a(1+r+r^2)\big)=700$

$\implies ar = 10 \\ \\ \text{So:}\ a = \frac{10}{r} , ar= 10 , ar^2 = 10r \\ \\ \frac{10}{r}+10+10r = 35 = 5 + 10 + 20 \\ \\ \text{So by inspection we get: r = 2},\frac{1}2 \\ \\ \text{\bold{Therefore the numbers in GP satisfying the following conditions are} :} \\ \\ \boxed{5,10,20}$

Hope this helps.