Question

How many terms of the series 230+227+224+....make a sum of 4200?

Answer 1

According to the question the given series is in arithmetic progression.

The first term in the series is 230 and the common difference is 3.

And the sum of n terms is given as 4200.

S = n/2[2a + (n - 1)d] = [a {a + (n - 1)d}] = [a + l]

4200 = n/2[2*230 + (n - 1)*-3]

or, 8400 = 460n - 3n² + 3n

or, 3n² - 463n + 8400 = 0

or, 3n² - (400+63)n + 8400 = 0

or, 3n² - 400n - 63n + 8400 = 0

or, n(3n - 400) - 21(3n - 400) = 0

or, n = 21 is the answer because count cannot be a fraction.

Answer 2

To find the terms AP,

Sn=n/2(2a+(n-1)d) now we have to assume the numbers such as a=230,

d= -3(difference between each numbers) and AP=4200.

Apply the formula

4200=n/2(2(230)+(n-1)-3),

=>8400=n(460-3n+3),

=> 8400=463n-3n^2 => 3n^2-463n+8400=0,

=>3n^2-(400+63)n+8400=0

therefore the overall n value is of 21 even it comes fraction take approximately n value.