Question

$\huge\boxed{\fcolorbox{pink}{silver}{Question}}$
Find 3 numbers in G.P. such that their sum is 21 and sum of squares is 189.
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Answer 1

Your answer is in the Attachment.

I hope this helps you!

Answer 2

Answer:

YOUR QUESTION IS ::-Sum of three terms which are in G.P. is 21. Sum of the square of the terms is 189. Find the terms.

AND ANSWER

Let the three numbers in G.P be a,ar,ar2

Since the sum of the numbers is 21

∴a+ar+ar2=21

⇒a(1+r+r2)=21 

Since the sum of the squares of the numbers is 189

∴a2+a2r2+a2r4=189

⇒a2(1+r2+r4)=189     .

Squaring both sides of (1),

we get a2(1+r+r2)2=441      .......(3)

Dividing (3) by (2), we get

a2(1+r2+r4)a2(1+r+r2)2=189442

(1+r+r2)2=189441

⇒(1+r2+r4)(1+r+r2)2=921

⇒(1+2r2+r4−r2)(1+r+r2)2=37

⇒(r2+1)2−r2(1+r+r2)2=37

⇒(r2+1−r)(r2+1+r)(1+r+r2)2=

⇒r=2×25±52−4×2×2

=45±25−16

=45±9=45±3=48,42=2,21

When r=2 from (1)

a(1+2+22)=21⇒a=3

∴ Numbers are 3,3×2,3×22 or 3,6,12

When r=21 from (1)

a[1+21+41]=21

⇒a[44+2+1]=21

⇒a=3 b=6 c =12