How many terms of the series 3+8+13+18+........must be taken so that their sum is 1010?
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a = 3, d = 5, sum = 1010
We know that a.p = n/2(2a+(n-1)d)
1010=n/2(2 * 3 +(n-1) * 5)
1010=n/2(6 + 5n - 5)
1010 = n/2(5n + 1)
2020=5n^2+n
5n^2+n-2020=0
5n^2+101n - 100n - 2020=0
n(5n+101)-20(5n+101)=0
(n-20)(5n+101)=0
n=20.
Hope this helps!
We know that a.p = n/2(2a+(n-1)d)
1010=n/2(2 * 3 +(n-1) * 5)
1010=n/2(6 + 5n - 5)
1010 = n/2(5n + 1)
2020=5n^2+n
5n^2+n-2020=0
5n^2+101n - 100n - 2020=0
n(5n+101)-20(5n+101)=0
(n-20)(5n+101)=0
n=20.
Hope this helps!
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