Question

In an A.P : 25 , 22 , 19........ The Sn = 116 then find last term. L Arithmetic Professions

Answer 1

Heya !
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★Arithmetic Progressions ★
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→ Given A.P = 25 , 22 , 19.........


• First term a = 25 

• Common Difference d = 22 - 25 = -3 

• Sn = 116 


→last term l = ? 


Now we have the formula for the sum of n terms ,

=> Sn = n/2 [ 2a + ( n - 1 ) d ] 

=> 116 = n/2 [ 2 × 25 + ( n - 1 ) -3 ]

=> 232 = n [ 50 - 3n + 3 ]

=> 232 = n [ 53 - 3n ]


=> We have the Quadratic Equation : 


→ 3n² - 53n - 232 = 0


Now I am solving it further using the Quadratic formula . You can opt any method.
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•Here a = 3 , b = -53 and c= -232 

We have d = b² - 4ac 

=> d = (53)² - 4 × 3 × 232 

=> d = 2809 - 2784 


•°• d = 25 


Then we have x = - b +- √d/2a 


We will get two values here: 


=) x = 53 + 5 / 6 ( neglect this value as it comes in fractions )

=) x = 53 - 5/6 = 48/6 = 8 ✔



°•° n = 8 
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=> Now we have to find the last term . There are two ways : 


→First method= : Sn = n/2 ( a + l )


→ 116 = 8/2 ( 25 + l )

=> 116 = 100 + 4l

=> 4l = 16 


•°• Last term = 4 ✔ 


→Second Method : an = a + ( n - 1 )d 

=> an = 25 + 7 × -3 

=> an = 25 - 21 


•°• Last term = 4 ✔ 


★The last term of the A.P is 4 !!


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Answer 2

In an A.P : 25 , 22 , 19........ The Sn = 116 then find last term. L Arithmetic Professions