Question

Divide 32 into four parts which are the four terms of an AP such that the product of the first and the fourth terms is to the product of the second and the third terms as 7:15.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that divide 32 in to 4 parts which are in AP.

Let assume that the 4 parts in AP are a - 3d, a - d, a + d, a + 3d.

So,

$\rm \: a - 3d + a - d + a + d + a + 3d = 32$

$\rm \: 4a = 32$

$\color{green}\rm\implies \:\boxed{ \rm{ \:a = 8 \: }}$

Further given that, the product of the first and the fourth terms is to the product of the second and the third terms as 7:15.

$\rm \: \dfrac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \dfrac{7}{15}$

$\rm \: \dfrac{ {a}^{2} - {9d}^{2}}{ {a}^{2} - {d}^{2} } = \dfrac{7}{15}$

$\rm \: 15( {a}^{2} - {9d}^{2}) = 7( {a}^{2} - {d}^{2})$

$\rm \: 15 {a}^{2} - {135d}^{2}= 7{a}^{2} - 7{d}^{2}$

$\rm \: 15 {a}^{2} - {7a}^{2}= 135{d}^{2} - 7{d}^{2}$

$\rm \: {8a}^{2} = {128d}^{2}$

$\rm \: {a}^{2} = {16d}^{2}$

On substituting the value of a = 8, we get

$\rm \: {8}^{2} = {16d}^{2}$

$\rm \: {16d}^{2} = 64$

$\rm \: {d}^{2} = 4$

$\rm\implies \:d \: = \: \pm \: 2$

So, two cases arise

Case :- 1 When a = 8, d = 2

So, numbers are 2, 6, 10, 14

Case :- 2 When a = 8, d = - 2

So, numbers are 14, 10, 6, 2

$\rule{190pt}{2pt}$

Additional Information :-

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Answer 2

❥︎ANSWER

Let the four parts be (a−3d),(a−d),(a+d) and (a+3d). 

Then, Sum of the numbers =32

⟹(a−3d)+(a−d)+(a+d)+(a+3d)=32⟹4a=32⟹a=8

It is given that

(a−d)(a+d)(a−3d)(a+3d)=157

⟹a2−d2a2−9d2=157

⟹64−d264−9d2=157⟹128d2=512⟹d2=4⟹d=±2

Thus, the four parts are a−3d,a−d,a+d and 3d, i.e. 2,6,10,14.

❥︎THANKS

I hope it will help you.