Respuesta :
The four consecutive terms are 2,6,10,14.
Explanation:
Let the four consecutive terms of an arithmetic progression be (a - 3d), (a - d),(a + d), (a + 3d)
It is given that the sum of four consecutive terms is 32.
Thus, adding all the four consecutive terms, we have,
[tex]\begin{array}{r}{a-3 d+a-d+a+d+a+3 d=32} \\{4 a=32} \\{a=8}\end{array}[/tex]
Thus, the value of a is 8.
It is also given that the ratio of the product of first and last term to the product of the two middle terms is 7:15. Thud, we have,
[tex]\frac{(a-3d)(a+3d)}{(a-d)(a+d)} =\frac{7}{15}[/tex]
Simplifying the terms, we have,
[tex]$\begin{aligned} \frac{a^{2}-9 d^{2}}{a^{2}-d^{2}} &=\frac{7}{15} \\ 15\left(a^{2}-9 d^{2}\right) &=7\left(a^{2}-d^{2}\right) \\ 15 a^{2}-135 d^{2} &=7 a^{2}-7 d^{2} \\ 8 a^{2} &=128 d^{2} \end{aligned}$[/tex]
Now, substituting a = 8, we have,
[tex]$\begin{aligned} 8(64) &=128 d^{2} \\ 512 &=128 d^{2} \\ 4 &=d^{2} \\ 2 &=d \end{aligned}$[/tex]
Thus, the value of d is 2.
Now, we shall substitute the value of a and d in the terms (a - 3d), (a - d),(a + d), (a + 3d), we get, the four consecutive terms.
Thus, the four consecutive terms are 2,6,10,14.
