How many terms of the AP: 63,60,57.... should be taken so that their sum is 693
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a=63
D=60-63=-3
sm=693
by putting the formula of sum sm=n/2(2a+(n-1)D).
D=60-63=-3
sm=693
by putting the formula of sum sm=n/2(2a+(n-1)D).
Answered by
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difference=-3
a1=63
sum of ap=n/2[2*a1[n-1]d]
n/2[2*63[n-1]-3]=693
n[126-3n+3]=693*2
129n-3n^2=1386
-3n^2+129n-1386=0
now it is in quadratic equation so
a=-3,b=129,c=-1386
roots=-b plus of minus √b^2-4ac/2a
-129 + or - √129^2-4*-3*-1386/2*-3
-129 + or -√16641-16632/-6
-129+ or- √9/-6
-129-3/-6 or -129+3/-6
-132/-6 or -129+3/-6
22 or 21 are the terms should be taken
Hence 22 term is 0
a1=63
sum of ap=n/2[2*a1[n-1]d]
n/2[2*63[n-1]-3]=693
n[126-3n+3]=693*2
129n-3n^2=1386
-3n^2+129n-1386=0
now it is in quadratic equation so
a=-3,b=129,c=-1386
roots=-b plus of minus √b^2-4ac/2a
-129 + or - √129^2-4*-3*-1386/2*-3
-129 + or -√16641-16632/-6
-129+ or- √9/-6
-129-3/-6 or -129+3/-6
-132/-6 or -129+3/-6
22 or 21 are the terms should be taken
Hence 22 term is 0
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