Question

the sum of four consecutive numbers in an ap is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7 ratio 15 find the numbers

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 = 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 = 128d²
Putting the value of a = 8 in above we get
.8(8)² = 128d²128d² = 512d² = 512/128d²
= 4d = 2
So, the four consecutive numbers are 8 - (3*2)8 - 6 = 28 - 2 = 68 + 2 = 108 + (3*2)8 + 6 = 14
Four consecutive numbers are 2, 6, 10 and 14

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Answer 2

Answer is attached.

Here's is Answer key for whole maths Leaked board paper 2018 =>
brainly.in/question/3121329