Question

If a1,a2,a3 are in gp such that a1 + a2 + a3 =13 and (a1)^2 +(a2)^2 +(a3)^2 = 91 then find a and r ​

Answer 1

Given :-

• a1,a2,a3 are in GP.
• a1 + a2 + a3 =13
• (a1)^2 +(a2)^2 +(a3)^2 = 91

To find :-

• a & r ?

Formula used :-

• (a + b +c)² = a² + b² + c² + 2(ab + bc + ca)
• nth term of a GP series is Tn = ar^(n-1)

Solution :-

→ (a1 + a2 + a3) = 13 ------ Equation (1)

Squaring both sides we get,

→ (a1 + a2 + a3)² = 13²

→ a1² + a2² + a3² + 2(a1a2 + a2a3 + a3a1) = 169

→ 91 + 2(a1a2 + a2a3 + a3a1) = 169

→ (a1a2 + a2a3 + a3a1) = (169 - 91)/2

→ (a1a2 + a2a3 + a3a1) = (78/2) = 39 . ---- Equation (2)

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Now,

→ a1 = a

→ a2 = ar^(2-1) = ar

→ a3 = ar^(3-1) = ar²

Putting These value in Equation (1) & (2) we get,

→ (a + ar + ar²) = 13 ------- Equation (3)

and,

→ (a*ar + ar*ar² + ar²*a) = 39

→ ar(a + ar² + ar) = 39 ------- Equation (4).

__________________

Dividing Equation (4) by Equation (3) now, we get,

→ [ ar(a + ar + ar²) ] / [ (a + ar + ar²) ] = 39/13

→ ar = 3 .

Hence, if we Assume a = 1 ,

Than,

→ r = 3 .

and, now , if we check, Equation (1),

→ a + ar + ar² = 13

→ 1 + 1*3 + 1*3² = 13

→ 1 + 3 + 9 = 13

→ 13 = 13 (Satisfy).

Also,

→ (a)² + (ar)² + (ar²)² = 91

→ 1² + (3)² + (1*3²)² = 91

→ 1 + 9 + 9² = 91

→ 10 + 81 = 91

→ 91 = 91 (Satisfy).

Therefore, we can conclude That, value of a is 1 and r is 3.

Answer 2

......$\tt{\huge{\purple{ ................. }}}$QUESTION : If a1,a2,a3 are in gp such that a1 + a2 + a3 =13 and (a1)^2 +(a2)^2 +(a3)^2 = 91 then find a and r .$a_{1},l \: a_{2} \: and \: a_{3} \: are \: In \: G.P.$$a_{1} + a_{2} + a_{3} = 13$${a_{1}} ^ 2 + { a_{2} } ^ 2 + { a_{3} } ^ 2 = 91$$Let \\ \\ a_{1} = a \\ \\ a_{2} = b \\ \\ a_{3} = c$$a +b+c = 13$${ a } ^ 2 + { b } ^ 2 + { c } ^ 2 = 91$Solving the above equations, we get : $ab + bc + ca = 39$$Now \: : \\ \\ a = a \\ \\ b = a { r } ^ 1 \\ \\ c = a { r } ^ 2$Substuting ,$ar ( a + ar + ar^2 ) = 39$Substuting the value of : a1 + a2 + a3 =13$ar = 3$Now we need to solve by trial and error method.Substitute different values of a and corresponding values of r and see which value holds.Solving we get : $a = 1 \\ \\ r = 3 \: ......... ( A )$