Question

the sum of four consecutive numbers in AP is 37 and the ratio of the product of first and the last term is 7:15 find the number

Answer 1

Let (a - 3d) , (a - d) , (a + d) and (a + 3d) are four consecutive terms in AP.
Now, A/C to question,
Sum of four consecutive terms = 37
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 37
4a = 37 ⇒a = 37/4

Now, question again said , ratio of products of first and last term to the product of middle is 7 : 15
e.g., (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 ⇒a² = 16d²
Taking square root both sides,
a = ± 4d
so, d = ±a/4
so, d = ± 37/16

Now, numbers are :
(a - 3d) = 37/4 - 3×37/16 = 37/16
(a - d) = 37/4 - 37/16 = 111/16
(a + d) = 37/4 + 37/16 = 185/16
(a + d) = 37/16 + 3 ×37/16 = 259/16


[Note :- you can solve this question by assuming four consecutive terms a , a + d , a + 2d , a + 3d . I assume four terms a -3d , a - d , a + d , a + 3d because it makes easy to solve this type of questions . You can see how after adding whole terms we get a = 37/4 , but when you take first one then you have two variable outcomes . Like 4a + 6d = 37 . It hard to proceed , well you should try both the method :)]

Answer 2

Solution :

Let (a-3d),(a-d),(a+d),(a+2d) are

Four consecutive terms in A.P

According to the problem given,

Sum of four terms = 37

=> a-3d+a-d+a+d+a+3d=37

=> 4a = 37

=> a = 37/4 --( 1 )

ii ) Ratio product of first and last terms and middle

terms

= 7 : 15

=> (a-3d)(a+3d):(a-d)(a+d) = 7:15

=> [a²-(3d)²]:[a²-d²] = 7:15

=> (a²-9d²):(a²-d²)=7:15

=> 15(a²-9d²)=7(a²-d²)

=> 15a²-135d² = 7a²-7d²

=> -135d²+7d²=7a²-15a²

=> -128d² = - 8a²

=> d² = 8a²/128

=> d² = a²/16

=> d² = ( 37/4)²/16

=> d² = 37²/16²

d = ± 37/16

Therefore ,

i ) if a = 37/4 , d = 37/16

Required 4 terms are

a-3d = 37/4-3(37/16)=37/16

a-d= 37/4-37/16=111/16

a+d = 37/4+37/16=185/16

a+3d=37/4+3(37/16)=259/16

ii ) if a = 37/4 , d = -37/16

required 4 terms are

259/16,184/16,111/16,37/16

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