Question

How many tem of series 54,51,48 be taken so thier sum is 513?

Answer 1

Hi ,

54 , 51 , 48 , ... is in Arithmetic

progression

First term = a = 54 ,

Common difference = d = a2 - a1

d = 51 - 54 = -3

Let the sum of ' n ' terms = Sn = 513

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We know that ,

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

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n/2 [ 2× 54 + ( n - 1 ) ( -3 ) ] = 513

n [ 108 - 3n + 3 ] = 1026

n( 111 - 3n ) = 1026

111n -3n² = 1026

3n² - 111n + 1026 = 0

3n² - 54n - 57n + 18 × 57 = 0

3n( n - 18 ) - 57 ( n - 18 ) = 0

( n - 18 ) ( 3n - 57 ) = 0

n - 18 = 0 or 3n - 57 = 0

Therefore ,

n = 18 or n = 57/3 = 19

Sum of 18 terms or

Sum of 19 terms in the given A.P

= 513

I hope this helps you.

:)