Question

four consecutive numbers in an ap is 32 and the ratio of the last and the middle terms is 7:15 find the number

Answer 1

Let the number be
(a - 3d), (a - d), (a + d) & (a + 3d).
Sum = 32
a - 3d + a - d + a + d + a + 3d = 32
$4a = 32 \\ a = \frac{32}{4} = 8$
$\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15}$
Simplifying it, you will get:-
$8 {a}^{2} = 128 {d}^{2} \\ 8 {(8)}^{2} = 128 {d}^{2} \\ 128 {d}^{2} = 512 \\ {d}^{2} = \frac{512}{128} = 4 \\ d = \sqrt{4} = 2$
Therefore, the numbers are:-
(a - 3d) = (8 - 3 × 2) = 8 - 6 = 2
(a - d) = (8 - 2) = 6
(a + d) = (8 + 2) = 10
(a + 3d) = (8 + 6) = 14

The A.P. is 2, 6, 10, 14.

Well, I am also in class 10. This question came yesterday in the exam.