Question

Find the four numbers forming a G.P if the second number exceeds
the first by 21 and the fourth number is 336 more than the third.

Answer 1

$\textbf{Given:}$

$\textsf{In a G.P, Second term exceeds the first term by 21 and}$

$\textsf{fourth number is 336 more than the third}$

$\textbf{To find:}$

$\textsf{The G.P}$

$\textbf{Solution:}$

$\textbf{Formula\;used:}$

$\mathsf{The\;n\,th\;term\;of\;G.P,\;\;a,ar,ar^2,.\,.\,.\;is}$

$\boxed{\mathsf{t_n=ar^{n-1}}}$

$\mathsf{Let\;the\;four\;numbers\;in\;G.P\;be\;a,\;ar,\;ar^2,\;ar^3}$

$\textsf{As per given data,}$

$\mathsf{t_2-t_1=21}$

$\implies\mathsf{ar-a=21}$------(1)

$\mathsf{Also,}$

$\mathsf{t_4-t_3=336}$

$\implies\mathsf{ar^3-ar^2=336}$

$\implies\mathsf{r^2(ar-a)=336}$

$\implies\mathsf{r^2(21)=336}$

$\implies\mathsf{r^2=\dfrac{336}{21}}$

$\implies\mathsf{r^2=16}$

$\implies\mathsf{r=4}$

$\mathsf{From\;(1)}$

$\mathsf{4a-a=21}$

$\mathsf{3a=21}$

$\mathsf{a=\dfrac{21}{3}}$

$\implies\boxed{\mathsf{a=7}}$

$\therefore\textsf{The required numbers are}$

$\mathsf{7,\;28,\;112,\;448}$

Answer 2

Given : the four numbers forming a G.P

the second number exceeds

the first by 21 and the fourth number is 336 more than the third

To Find : four numbers forming a G.P

Solution:

four numbers forming a G.P

a , ar , ar² , ar³

second number exceeds the first by 21

=> ar - a  = 21

=> a(r - 1)  = 21

fourth number is 336 more than the third.

=> ar³ - ar² = 336

=> ar²(r - 1) = 336

=> ar²(r - 1) /  a(r - 1)  = 336/21

=>  r² = 16

=> r = ± 4

Case 1 :  r = 4

  a(r - 1)  = 21

=> a(4 - 1) = 21

=> a = 7

Numbers are   7  ,  28  , 112   , 448

Case 2 :  r = -4

  a(r - 1)  = 21

=> a(-4 - 1) = 21

=> a = -21/5

Numbers are   -21/5  ,  84/5    , -336/5   , 1344/5

four numbers are :

 7  ,  28  , 112   , 448    or

-21/5  ,  84/5    , -336/5   , 1344/5

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