Question

the sum of first n term of progression -30,-24,-18,-12.. is 540. find the value of n

Answer 1

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Answer 2

The value of n is 20.

Given:

A.P. : -30, -24, -18, -12, and so on

The sum of n terms=540

To find:

The value of n

Solution:

The first term of the given A.P., a= -30

The common difference, d= -24-(-30)= -24+30=6

The number of terms=n

The sum of the given A.P.=n/2×(2a+(n-1)d)

Using the values,

540=n/2×(2(-30)+(n-1)6)

540=n/2×(-60+6(n-1))

540= -60n/2+6n(n-1)/2

540= -30n+3n(n-1)

540= -30n+3$n^{2}$-3n

540= -33n+3$n^{2}$

3$n^{2}$-33n-540=0

$n^{2}$-11n-180=0

$n^{2}$-20n+9n-180=0

n(n-20)+9(n-20)=0

(n+9)(n-20)=0

So, n+9=0 and n-20=0

n= -9 and n=20.

Since n cannot be negative, n=20.

Therefore, the value of n is 20.