Question

how many terms of ap 1/2,1,3/2,2,5/2,3,7/2 are required to get a sum of 637.5​

Answer 1

Given : 1/2   ,  1 ,  3/2  ,  2 , 5/2  , 3  , 7/2

To Find : how many terms of ap are required to get a sum of 637.5​

Solution:

a = 1/2

d = 1/2

n = ?

S = (n/2)(2a + (n - 1)d)  = 637.5​

=> n ( 1 +  (n - 1) (1/2)  ) =  1275

=> n (  2 + n - 1)  = 2550

=> n ( n + 1)  = 2550

=>  n ( n + 1)  = 50(51)

=> n = 50

Hence 50 terms of AP are required

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Answer 2

Given :- how many terms of ap 1/2,1,3/2,2,5/2,3,7/2 are required to get a sum of 637.5 .

Solution :-

Let n terms of AP makes the sum as 637.5 .

so,

→ First term = a = 1/2

→ Common difference = d = 1 - 1/2 = 1/2

→ nth term = n

then,

→ Sn = (n/2)[2a + (n - 1)d]

→ 637.5 = (n/2)[2 * (1/2) + (n - 1)(1/2)]

→ 1275 = n[1 + 0.5n - 0.5]

→ 1275 = 0.5n² + 0.5n

→ 0.5n² + 0.5n - 1275 = 0

→ n² + n - 2550 = 0

→ n² + 51n - 50n - 2550 = 0

→ n(n + 51) - 50(n + 51) = 0

→ (n + 51)(n - 50) = 0

→ n = (-51) and 50 .

since negative value of nth term is not possible .

therefore, value of n is equal to 50 .

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