Question

The sum of three consecutive terms in an arithmetic series is 36 and their multiplication is 1140; then find those terms. (The terms are in descending order.) ​

Answer 1

Answer:

Given:

• Sum of three consecutive terms in an arithmetic series is 36
• Their multiplication is 1140

Find:

• The terms?

Solution:

Let us take the terms as a - d, a, a + d

From question, their sum is 36

$\to{ \sf{a - d + a + a + d = 36}} \\ \\ \to{ \sf{3a \ = 36}} \\ \\ \to{ \sf{a = \frac{36}{3} }} \\ \\ \to{ \sf{a = 12}}$

So, the value of a is 12

From question, their multiplication is 1140

$\to{ \sf{(a - d)(a)(a + d) = 1140}} \\ \\ \to{ \sf{(a)( { a}^{2} - {d}^{2} ) = 1140}} \\ \\ \to{ \sf{(12)( {12}^{2} - {d}^{2} ) = 1140}} \\ \\ \to{ \sf{(12)(144 - {d}^{2}) = 1140 }} \\ \\ \to{ \sf{(144 - {d}^{2} ) = \frac{1140}{12} }} \\ \\ \to{ \sf{(144 - {d}^{2}) = 95 }} \\ \\ \to{ \sf{ { - d}^{2} = 95 - 144}} \\ \\ \to{ \sf{ {d}^{2} = 49 }} \\ \\ \to{ \sf{d = \sqrt{49} }} \\ \\ \to{ \sf{d = 7}}$

so, the value of d is 7

By substituting these values :

⇒ a - d = 12 - 7 = 5

⇒ a = 12

⇒ a + d = 12 + 7 = 19

Therefore,

• The terms in descending order are 19, 12, 5

Thank you....!

Answer 2

Given : The sum of three consecutive terms in an arithmetic series

36 and their multiplication 1140, then find those terms.

To Find : Terms  in descending order

Solution:

Terms

a- d  , a  , a  + d

a - d +  a  + a + d  = 36

=> 3a  = 36

=>  a = 12

(12 - d) , 12 , ( 12 + d)

(12 - d)12(12 + d)  = 1140

=> 144 - d² =  95

=> d² = 49

=> d = ±7

a = 12  ,  d = ±7

a- d  , a  , a  + d

=> 5  , 12  , 19

or  19 , 12 , 5

Terms  in descending order  19 , 12 , 5

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