14) How many terms of the AP 20, 19 1/ 3, 18 2 / 3 must be taken so that their sum is 300 and Explain the double answer.
Answers
HEY THERE!!!
◀Given arithmetic sequence or progressions ;-
◀ Ap :- 20 , 58 / 3 , 56 / 3 ---- 300
Here,
◀ Given;-
first term=20
common difference= 56 / 3 - 20 = - 2 / 3
◀ We know that formula of summation;-
Sn = n / 2 ( 2a + (n-1)d )
◀ 300 = n / 2 [ 2(20) + (n-1)(- 2/3) ]
◀ 600 = n ( 40 - 2n / 3 + 2/3 )
◀ 600 x 3 = n ( 120 - 2n + 2 )
◀ 1800 = n ( 122 - 2n )
◀ 1800=122n-2n²
Arrange in Quadratic form;-
◀ 2n² - 122n + 1800 = 0
Taken common 2 from Quadratic form;-
→ n² - 61n + 900 = 0
→n² - 36n - 25n + 900 = 0
→ n ( n - 36 ) - 25 ( n - 36 ) = 0
→ ( n - 25 ) ( n - 36 ) = 0
→ n = 25 or 36 .
Hence, Two Sum of number obtained = 300
→ Solving both Equaton;-
→ S25 = 25 / 2 ( 40 + (25 - 1) ( - 2 / 3 ) = 25 / 2 ( 40 - 16 )
→ 25 / 2 ( 24 )
→ 25 x 12 = 300
→ S36 = 36 / 2 ( 40 + (36 - 1) ( - 2 / 3 )
→ 18 ( 40 + 35 ( - 2 / 3 )
→ 18 ( 40 - 70 / 3 )
→18 x 50 / 3
→ 6 x 50
→ 300.
Hence, Required numbers=25 and 36
Answer:
- Required numbers=25 and 36
Step-by-step explanation:
Given arithmetic sequence or progressions:-
AP:- 20 , 58 / 3 , 56 / 3 ---- 300
Here,
Given:-
First term=20
Common difference= 56 / 3 - 20 = - 2 / 3
We know that formula of summation:-
Sn = n / 2 ( 2a + (n-1)d )
300 = n / 2 [ 2(20) + (n-1)(- 2/3) ]
600 = n ( 40 - 2n / 3 + 2/3 )
600 x 3 = n ( 120 - 2n + 2 )
1800 = n ( 122 - 2n )
1800=122n-2n²
Arrange in Quadratic form:-
2n² - 122n + 1800 = 0
Taken common 2 from Quadratic form:-
→ n² - 61n + 900 = 0
→n² - 36n - 25n + 900 = 0
→ n ( n - 36 ) - 25 ( n - 36 ) = 0
→ ( n - 25 ) ( n - 36 ) = 0
→ n = 25 or 36 .
Hence, Two Sum of number obtained = 300
Solving both Equation:-
→ S25 = 25 / 2 ( 40 + (25 - 1) ( - 2 / 3 ) = 25 / 2 ( 40 - 16 )
→ 25 / 2 ( 24 )
→ 25 x 12 = 300
→ S36 = 36 / 2 ( 40 + (36 - 1) ( - 2 / 3 )
→ 18 ( 40 + 35 ( - 2 / 3 )
→ 18 ( 40 - 70 / 3 )
→18 x 50 / 3
→ 6 x 50
→ 300.
Hence, Required numbers=25 and 36