If 13,?x,?y, and 31 are the four consecutive terms of an A.P., then what are the respective values of?x?and?y?

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Solution:- Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d) So, according to the question. a-3d + a - d + a + d + a + 3d = 32 4a = 32 a = 32/4 a = 8 ......(1) Now, (a - 3d)(a + 3d)/(a - d)(a + d) = 7/15 15(a² - 9d²) = 7(a² - d²) 15a² - 135d² = 7a² - 7d² 15a² - 7a² = 135d² - 7d²  8a² = 128d² Putting the value of a = 8 in above we get. 8(8)² = 128d² 128d² = 512 d² = 512/128