How many terms of A. P. 63,60,57,.......taken so that their sum is 693?
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Answered by
18
Hey!
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Given A.P = 63 , 60 , 57 ....
Here,
a = 63
d = 60 - 63 = -3
We know,
S = n/2 [ 2a + (n - 1) d ]
693 = n/2 [ 2×63 + (n - 1) -3 ]
693 = n/2 ( 126 - 3n + 3 )
693 × 2 = n ( 129 - 3n )
1386 = n ( 129 - 3n)
1386 = 129n - 3n²
1386 - 129n +3n²
Factorising -:
3n² - 66n - 63n + 1386 = 0
(3n² - 66n) - (63n - 1386) = 0
3n (n - 22) - 63 ( n - 22) = 0
(3n -63) (n - 22) = 0
So,
3n - 63 = 0
n = 63/3
n = 21
Or
n - 22 = 0
n = 22
_____________
Hope it helps...!!!
_____________
Given A.P = 63 , 60 , 57 ....
Here,
a = 63
d = 60 - 63 = -3
We know,
S = n/2 [ 2a + (n - 1) d ]
693 = n/2 [ 2×63 + (n - 1) -3 ]
693 = n/2 ( 126 - 3n + 3 )
693 × 2 = n ( 129 - 3n )
1386 = n ( 129 - 3n)
1386 = 129n - 3n²
1386 - 129n +3n²
Factorising -:
3n² - 66n - 63n + 1386 = 0
(3n² - 66n) - (63n - 1386) = 0
3n (n - 22) - 63 ( n - 22) = 0
(3n -63) (n - 22) = 0
So,
3n - 63 = 0
n = 63/3
n = 21
Or
n - 22 = 0
n = 22
_____________
Hope it helps...!!!
jay376:
very very nice
Answered by
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