Question

⭐How many terms of the A. P 54,51,48......Should be taken so that their sum is 513?⭐

CLASS 10 CHAPTER :ARITHMETIC PROGRESSIONS

BEST ANSWER WILL BE MARKED AS BRAINLIEST!

Answer 1

$\huge{\mathfrak{heya!!}}$

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Solution:-
>>>>>>>>>>>>>>>>>>>>>>>>>>

Given :- AP is 54,51,48....

sum =513
here, a = 54, d = 51-54=-3

Let the number of terms be
$n$
Sum of n terms of AP =
$\frac{n}{2} (2a + (n - 1)d)$

sum of n terms of Ap = n/2 ( 2×54+(n-1) -3)

513 =n/2 ( 2×54+(n-1) -3)
..
513= n/2(108-3n+3)

2× 513=108n-3n^2+3n

1026=-3n^2+111n

3n^2-111n+1026=0

3n^2-57n-54n+1026=0

(3n-54)(n-19)=0

so, n = 19 or n=54/3=18.

Thanks