Question

how many terms of 54,51,48...be taken so that sum is 5/3

Answer 1

Given series is 54,51, 48,....54,51,48,....

s_n=\dfrac{n}{2}[2a+(n-1)d]s

n

=

2

n

[2a+(n−1)d]

\therefore 513=\dfrac{n}{2}[2\times 54+(n-1)-3∴513=

2

n

[2×54+(n−1)−3

1026=n[108-3n+3]1026=n[108−3n+3]

1026=111n-3n^21026=111n−3n

2

3n^2-111n+1026=03n

2

−111n+1026=0

n^2-37n+342=0n

2

−37n+342=0

n^2-18n-19n+342=0n

2

−18n−19n+342=0

n(n-18)-19(n-18)=0n(n−18)−19(n−18)=0

(n-18)(n-19)=0(n−18)(n−19)=0

Then n=18n=18 and n=19n=19

hope this helps

maek me as Brainliest