Divide 56 into 4 parts which are in AP such that hte ratio of product of extremes to the product - Maths - Arithmetic Progressions

Question

Divide 56 into 4 parts which are in AP such that hte ratio of
product of extremes to the product of mean is 5 : 6? Hoping for a
quick response plz [ May be within Sunday ]

Harleen Birdi · Accepted Answer

Dear Student! Let the four parts be (a â€“ 3d), (a â€“ d), (a + d) and (a + 3d) Then, sum = 56 â‡’ (a â€“ 3d) + (a â€“ d) + (a + d) + (a + 3d) = 56 â‡’ 4a = 56 â‡’ a = 14 It is given that $\frac{(a - 3 d) (a + 3 d)}{(a - d) (a + d)} = \frac{5}{6} \Rightarrow \frac{a^{2} - 9 d^{2}}{a^{2} - d^{2}} = \frac{5}{6} \Rightarrow \frac{(14)^{2} - 9 d^{2}}{(14)^{2} - 9 d^{2}} = \frac{5}{6} (Putting a = 14) \Rightarrow \frac{196 - 9 d^{2}}{196 - d^{2}} = \frac{5}{6}$ $\Rightarrow 1176 - 54 d^{2} = 980 - 5 d^{2} \Rightarrow 1176 - 980 = 54 d^{2} - 5 d^{2} \Rightarrow 49 d^{2} = 196 \Rightarrow d^{2} = 4 \Rightarrow d = \pm 2$ Thus, the four parts are a â€“ 3d, a â€“ d, a + d, a + 3d i e , 8, 12, 16 and 20 Cheers !

Divide 56 into 4 parts which are in AP such that hte ratio of product of extremes to the product of mean is 5 : 6? Hoping for a quick response......plz.....[ May be within Sunday ]