How many terms of the arithmetic progression 63,60,57..........must be taken so that their sum is 693?
Answers
Answered by
2
first term a = 63
common difference , d = (-3)
sum = n/2[2a+(n-1)d]
693 = n/2[2*63+(n-1)(-3)]
1386 = n [126-3n+3]
1386 = 129n-3n²
3(n²-43n+462) = 0
n²-43n + 462 = 0
n²-22n-21n + 462 = 0
n(n-22)-21(n-22) = 0
(n-21)(n-22) = 0
n = 21 & 22
n = 22 answer
[ mark as brainlist if you understand ]
Similar questions