How many terms of AP.....65,60,55...... should be taken so that the sum is zero?
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Given A.P is 65,60,55.......
Here a = 65 and d = -5
Sum is 0⇒ s=0
s=n/2(2a+(n-1)d)
0 = n/2(2*65+(n-1)-5)
0 = n/2(130+5n-5)
0 = 125n+5n²
5n² = -125n
n² = -25
⇒ n = +5 or -5
-5 is not possible
So answer is 5
Here a = 65 and d = -5
Sum is 0⇒ s=0
s=n/2(2a+(n-1)d)
0 = n/2(2*65+(n-1)-5)
0 = n/2(130+5n-5)
0 = 125n+5n²
5n² = -125n
n² = -25
⇒ n = +5 or -5
-5 is not possible
So answer is 5
nishita19:
I think sum of 27 terms is zero
there is a mistake in 7 line....
so... Answer is wrong
but correct answer is 27
never mind please
Sorry
i know you are intelligent more than me
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12
Here Is Your Answer My Baby...!!!♥️♥️✌️✌️
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