Math, asked by spankaj1924ow82im, 1 year ago

How many terms of the series 3+8+13+18+........must be taken so that their sum is 1010?

Answers

Answered by siddhartharao77
7
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!
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