Question

can someone solve the question of 4marks AP Q28 maths 10th paper

Answer 1

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
d² = 4
d = 2
So, the four consecutive numbers are
8 - (3*2)
8 - 6 = 2
8 - 2 = 6
8 + 2 = 10
8 + (3*2)
8 + 6 = 14
Four consecutive numbers are 2, 6, 10 and 14

Answer 2

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 = 324a = 32a = 32/4a = 8 ......(1)Now, (a - 3d)(a + 3d)/(a - d)(a + d) = 7/1515(a² - 9d²) = 7(a² - d²)15a² - 135d² = 7a² - 7d²15a² - 7a² = 135d² - 7d² 8a² = 128d²Putting the value of a = 8 in above we get.8(8)² = 128d²128d² = 512d² = 512/128d² = 4d = 2So, the four consecutive numbers are 8 - (3*2)8 - 6 = 28 - 2 = 68 + 2 = 108 + (3*2)8 + 6 = 14Four consecutive numbers are 2, 6, 10 and 14