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Question: Divide 56 in four parts in AP, such that the ratio of the product of extremes to the product of mean...

Divide 56 in four parts in AP, such that the ratio of the product of extremes to the product of means is 5:6.

Explanation

Solution

Hint: Assume that the first term of the formed A.P is a and the common difference is d. Hence the number will be a, a+d,a+2d,a+3d. The sum of the terms of the A.P is 56, and the ratio of the product of extremes to the product of the means is 5:6. Using this information form, two equations in a and d. Solve for a and d. Hence find the partition. Alternatively, we can select four terms in A.P as a-3d, a-d, a+d, a+3d. Selecting terms in this way makes calculations much easier.

Complete step-by-step solution -
Let the first term of the A.P formed by the partition be a and the common difference be d.
Hence the numbers are a, a+d, a+2d and a+3d.
Now we have a+a+d+a+2d+a+3d = 56, i.e. 4a+6d = 56
Dividing both sides by 2, we get
2a+3d = 28 (i).
Also, the ratio of the product of extremes to the product of means is 5:6. Hence we have
(a)(a+3d)(a+d)(a+2d)=56\dfrac{\left( a \right)\left( a+3d \right)}{\left( a+d \right)\left( a+2d \right)}=\dfrac{5}{6}
Cross multiplying, we get
6a(a+3d)=5(a+d)(a+2d)6a\left( a+3d \right)=5\left( a+d \right)\left( a+2d \right)
Expanding the terms, we get
6a2+18ad=5a2+10ad+5ad+10d26{{a}^{2}}+18ad=5{{a}^{2}}+10ad+5ad+10{{d}^{2}}
Subtracting 5a25{{a}^{2}} from both sides, we get
a2+18ad=15ad+10d2{{a}^{2}}+18ad=15ad+10{{d}^{2}}
Subtracting 6ad from both sides, we get
a2=−3ad+10d2 (ii){{a}^{2}}=-3ad+10{{d}^{2}}\text{ (ii)}
From equation (i), we have a=28−3d2a=\dfrac{28-3d}{2}
Substituting the value of “a” in equation (ii), we get
(28−3d2)2=−3(28−3d2)d+10d2{{\left( \dfrac{28-3d}{2} \right)}^{2}}=-3\left( \dfrac{28-3d}{2} \right)d+10{{d}^{2}}
Using (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}, we get
196+94d2−42d=−42d+92d2+10d2196+\dfrac{9}{4}{{d}^{2}}-42d=-42d+\dfrac{9}{2}{{d}^{2}}+10{{d}^{2}}
Transposing all d2{{d}^{2}} terms on LHS
94d2−92d2−10d2+196−42d=−42d ⇒9−18−404d2+196−42d=−42d ⇒−494d2+196−42d=42d \begin{aligned} & \dfrac{9}{4}{{d}^{2}}-\dfrac{9}{2}{{d}^{2}}-10{{d}^{2}}+196-42d=-42d \\\ & \Rightarrow \dfrac{9-18-40}{4}{{d}^{2}}+196-42d=-42d \\\ & \Rightarrow \dfrac{-49}{4}{{d}^{2}}+196-42d=42d \\\ \end{aligned}
Adding 42d from both sides, we get
−494d2+196=0\dfrac{-49}{4}{{d}^{2}}+196=0
Multiplying LHS and RHS by 4, we get
−49d2+784=0-49{{d}^{2}}+784=0
Hence we have
49d2=78449{{d}^{2}}=784
Dividing both sides by 13, we get
d2=78449=16{{d}^{2}}=\dfrac{784}{49}=16
Hence d = 4 or -4
When d = 4, we have
2a+12=28
i.e 2a = 16
Hence a = 8.
Hence the terms are 8,12,16,20
When d = -4, we have
2a-12 = 28
i.e a = 20
Hence the terms are 20,16,12,8

Note: Alternative method: Best Method
Let the terms be a-3d, a-d, a+d and a+3d
Hence we have
a-3d+a-d+a+d+a+3d = 56
i.e. 4a = 56
Hence a=14
Also we have
(a−3d)(a+3d)(a−d)(a+d)=56 ⇒a2−9d2a2−d2=56 \begin{aligned} & \dfrac{\left( a-3d \right)\left( a+3d \right)}{\left( a-d \right)\left( a+d \right)}=\dfrac{5}{6} \\\ & \Rightarrow \dfrac{{{a}^{2}}-9{{d}^{2}}}{{{a}^{2}}-{{d}^{2}}}=\dfrac{5}{6} \\\ \end{aligned}
Cross multiplying and simplifying, we get
a2=49d2{{a}^{2}}=49{{d}^{2}}
Hence 7d=a,−a7d=a,-a
Since a = 14, we have
7d =14 or 7d = -14
Hence d = 2 or d = -2
Hence the terms are 8,12,16,20 or 20,16,12 and 8 which is the same as obtained above.