Question

three numbers are in AP and their sum is 21 if 1, 5, 15 are added to them respectively they form a GP then the numbers are​

Answer 1

Answer:

$\red {Required \: 3 \: terms \: of \: G.P \:are }$

$\green { \: 6, 12 \:and \: 24 }$

Step-by-step explanation:

$Let \: (a-d), a \:and\: (a+d) \: are \: in \: A.P$

$Sum \: of \: three \: terms = 21 \:(given)$

$\implies a-d+a+a+d = 21$

$\implies 3a = 21$

$\implies a = \frac{21}{3} = 7$

$Required \: 3 \:terms \:of \:A.P ,\:are \\(7-d),7, (7+d)$

/* According to the problem given,

$If \:we \: added \: to \: 1,5\: and \: 15 \: to \\above \:3 \:terms \: A.P ,\:we \:get \\3 \:terms \:of \: G.P. They \:are$

$First \:term (t_{1}) = 7-d+1 = 8-d$

$Second \:term (t_{2}) = 7+5 = 12$

$Third \:term (t_{3}) = 7+d+15 = 22+d$

$We \: know \: that , t^{2}_{2} = t_{1} \times t_{2}$

$\implies 12^{2} = (8-d)(22+d)$

$\implies 144 = 8d + 176 - d^{2} - 22d$

$\implies d^{2} + 14d + 144 - 176 = 0$

$\implies d^{2} + 14d - 32 = 0$

$\implies d^{2} + 16d - 2d - 32 = 0$

$\implies d( d + 16 ) - 2( d + 16 ) = 0$

$\implies (d+16)(d-2) = 0$

$\implies d + 16 = 0 \: Or \: d - 2 = 0$

$\implies d = - 16 \: Or \: d = 2$

$Required \: G.P : \\</p><p>\underline { Case \: 1}$

$If \: d = -16,$

$First \:term (t_{1}) = 8-(-16) = 8 + 16 = 24$

$Second \:term (t_{2}) = 12$

$Third \:term (t_{3}) = -16+22= 6$

$\underline { Case \: 2}$

$If \: d = 2,$

$First \:term (t_{1}) = 8-2= 6$

$Second \:term (t_{2}) = 12$

$Third \:term (t_{3}) = 2+22= 24$

•••♪

Answer 2

ANSWER : 5,7,9

STEP BY STEP EXPLANATION:

5+7+9=21

so if we add 1,5,15 to each number (i.e)

      5  7  9

(+)  1   5  15

     6  12  24

here by finding r that is r=12/6=2 and 24/12=2

therefore r=2

and it is in GP

therefore the numbers are 5,7,9