Question

How many terms of the ap 20, 19 1/3, should be taken so that their sum is 300

Answer 1

A = 20 d = - 2/3Sn = 300n/2 [2a + (n-1)d] = 300n/2 [2 * 20 = (n-1) – 2/3] =300n/2[40-2/3n+2/3]=300Proceed uwill get an eqnie n2-61n+900=0now solvethe eqn uwill get n as 25 or36this implies that the whole of 25 to 36 th term I.

Answer 2

Answer:

Step-by-step explanation:

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