the sum of three terms of gp is 13 and sum of their squares is 91 what are the three terms

Question

I tried to solve this question bt i  didn't.....please give the answer of this question

Answer 1

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Answer 2

Grace

Answer 3

GP three numbers a, ar , ar^2 Sum of three numbers = 13 a+ar+ar^2 = 13 a (1+r+r^2) = 13 Square on both sides a^2 (1+r+r^2)^2 = 13^2 a^2 ( 1+ r + r^2)^2 = 169 ------>. 1 Sum of their squares. = 91 a^2 + (ar)^2 + (ar^2)^2. = 91 a^2 + a^2*r^2 + a^2*r^4 = 91 a^2 ( 1 + r^2 + r^4 ) = 91 ----------->. 2 Divide equation from 2 to 1 (a^2 ( 1 + r^2 + r^4 ) ) / (a^2 ( 1+ r + r^2)^2) = 91 / 169 1+r^2 + r^4 / ( 1+ r + r^2 )^2 = 7/13 Now 1+r^2 +r^4 = 1+ 2r^2 + r^4 - r^2 (a +b )^2 = a^2 +2ab + b^2 = (1+r^2)^2 - r^2 a^2 - b^2 = (a +b ) (a - b) = ( 1 + r^2 + r) ( 1 + r^2 - r) Rearrange and write = ( 1 + r + r^2 ) (1 - r + r^2) ( 1 + r + r^2 ) ( 1 - r + r^2 ) / ( 1 + r + r^2)^2 = 7/13 (1- r + r^2 ) / (1 + r + r^2) =7/13 Cross multiply 13 ( 1 - r + r^2 ) = 7 ( 1+ r + r^2 ) 13 -13r +13r^2 = 7 +7r +7r^2 13 -13r + 13r^2 -7 -7r-7r^2 = 0 6 - 20r + 6r^2 = 0 2 ( 3 - 10r + 3r^2 ) = 0 3r^2 - 10r + 3 = 0/2 3r^2 -10r +3 = 0 Factorie the following 3r^2 -9r - 1r + 3 = 0 3r ( r - 3 ) -1 ( r - 3 ) = 0 ( r - 3 ) (3r -1) = 0 r = 3 and r = 1/3 Find a: r = 3 a ( 1+ r + r^2 ) = 13 a ( 1+ 3 + 3^2 ) = 13 a ( 1 + 3 + 9 ) =13 a (13) = 13 a = 13/13 a = 1 The first three numbers of GP a , ar , ar ^2 When. a = 1 and r = 3 1,3,9

Answer 4

Three terms can be represnted by a, ar, ar^2 where a is the first term and r the common ratio.

a(1+r+r^2)=13 so a^2(1+r+r^2)^2=169 by squaring.

a^2+a^r^2+a^2r^4=91=a^2(1+r^2+r^4)

Divide the two equations: 169/91=(1+r+r^2)^2/(1+r^2+r^4).

(r^3-1)=(r-1)(1+r+r^2) and (r^6-1)=(r^2-1)(1+r^2+r^4);

1+r+r^2=(r^3-1)/(r-1); 1+r^2+r^4=(r^6-1)/(r^2-1); (1+r+r^2)^2=(r^3-1)^2/(r-1)^2.

So 169/91=13/7=(r^3-1)^2(r^2-1)/((r-1)^2(r^6-1))=(r^3-1)(r+1)/((r^3+1)(r-1))=(r^3-1)/((r^2-r+1)(r-1))=(r^2+r+1)/(r^2-r+1).

Therefore cross-multiplying: 13(r^2-r+1)=7(r^2+r+1); 6r^2-20r+6=0; 3r^2-10r+3=0.

This factorises: (3r-1)(r-3)=0, giving us two complementary zeroes: r=3 and r=1/3.

So we can find a: a(1+r+r^2)=13; 13a=13, so a=1 when we substitute r=3.

The three terms are 1, 3, 9 because 1+3+9=13 and 1+9+81=91.

If we use r=1/3 we get 9, 3, 1 as the terms.