Question

How many terms of the AP 20 19 1/3 18 2/3 must be taken so that their sum

Answer 1

HEY THERE!!!
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◀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

Heya...

====== Solution =====

First term given = 20

Common difference = 56/3-20 = -2/3

Sn = n/2[2a+(n-1)d]

300 = n/2[2(20)+(n-1)(-2/3)]

600 = n(40-2n/3+2/3)

600*3=n(120-2n+2)

1800 = 122n-2n^2

Equation formed :-

2n^2-122n+1800 = 0

For solving n :-

n^2-61n+900 =0
n^2-36n-25+900 = 0

n(n-36)-25(n-36) =0

(n-25) (n-36) =0

n = 25 & 36

For sum of obtained numbers...

S25 = 25/2[40+(25-1)(-2/3)]
25/2*(24)
25*12 = 300

S36 = 36/2[40+(36-1)(-2/3)]
6*50 = 300 ....

So... 25 and 36 are required no..

Thank you