Question

how many terms of the AP 20, 19 1 upon 3, 18 2 upon 3 must be taken so that their sum is 300 and explain the double answer

Answer 1

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 2

Answer:

Step-by-step explanation: