Question

Thw sum of four consecutive terms in an



a.P is 32 and the ratio of the product of the first and last term to the product of 2 middle terms is 7:15 find the number

Answer 1

$\textsf{Let the consecutive terms be\ \ a,\ a+d,\ a+2d,\ a+3d .} \\ \\ \\ \textsf{Sum of the terms = \ 32 } \\ \\ a+(a+d)+(a+2d)+(a+3d)=32 \\ \\ a+a+d+a+2d+a+3d=32 \\ \\ 4a+6d=32 \\ \\ 4a=32-6d \\ \\ 2a=16-3d\ \ \ \ \ \longrightarrow\ \ \ (1)$

$\textsf{Product of first and fourth terms : Product of second and third terms = 7 : 15} \\ \\ \\ a(a+3d) : (a+d)(a+2d)=7:15 \\ \\ \\ a^2+3ad:a^2+3ad+2d^2=7:15 \\ \\ \\ \textsf{Taking \ a^2+3ad = k ,} \\ \\ \\ k:k+2d^2=7:15 \\ \\ \\ \displaystyle \frac{k}{k+2d^2}=\frac{7}{15} \\ \\ \\ 15k=7(k+2d^2) \\ \\ \\ 15k=7k+14d^2 \\ \\ \\ 15k-7k=14d^2 \\ \\ \\ 8k=14d^2$

$\textsf{Putting \ k = a^2+3ad ,} \\ \\ \\ 8(a^2+3ad)=14d^2 \\ \\ \\ 8a^2+24ad=14d^2 \\ \\ \\ 8a^2+24ad-14d^2=0 \\ \\ \\ 8a^2-4ad+28ad-14d^2=0 \\ \\ \\ 4a(2a-d)+14d(2a-d)=0 \\ \\ \\ (4a+14d)(2a-d)=0 \\ \\ \\ 2(2a+7d)(2a-d)=0 \\ \\ \\ (2a+7d)(2a-d)=0 \\ \\ \\ (16-3d+7d)(16-3d-d)=0\ \ \ \ \ \ \ \ \ \ \textsf{[From (1) ]} \\ \\ \\ (16+4d)(16-4d)=0$

$\begin{tabular}{|r|l|}\cline{1-2}16 + 4d = 0\ \ \ &\ \ \ 16 - 4d = 0 \\ \cline{1-2}4d = -16\ \ \ &\ \ \ 4d = 16 \\ \cline{1-2} d = \textbf{-4}\ \ \ &\ \ \ d=\textbf{4} \\ \cline{1-2} \end{tabular}$

$\therefore\ \ d=\pm4$

$\textsf{From\ (1) ,} \\ \\ \\ 2a=16-3d \\ \\ \\ 2a=16-3 \times \pm4 \\ \\ \\ 2a=16-(\pm12) \\ \\ \\ 2a=16\mp12 \\ \\ \\ 2a=4\ \ \ \ \ \ \ \ \ \ ; \ \ \ \ \ \ \ \ \ \ 2a=28 \\ \\ \\ a=\bold{2}\ \ \ \ \ \ \ \ \ \ \ ; \ \ \ \ \ \ \ \ \ \ \ \ a=\bold{14} \\ \\ \\ \textsf{[Note: The values of a are according to the corresponding values of d , i.e.,} \\ \textsf{ a=2 for d=4 and a=14 for d=-4 .]}$

$\textsf{On taking a=2\ \ \&\ \ d=4 ,} \\ \\ \\ a+d=\bold{6} \\ \\ a+2d=\bold{10} \\ \\ a+3d=\bold{14} \\ \\ \\ \\ \textsf{On taking a=14\ \ \&\ \ d=-4 ,} \\ \\ \\ a+d=\bold{10} \\ \\ a+2d=\bold{6} \\ \\ a+3d=\bold{2}$

$\Large \textsf{Whatever, the terms are}\ \ \textbf{2, 6, 10 \& 14}$