How many terms of the series 3+8+13+18+........must be taken so that their sum is 1010?
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hi friend,
the given series is in arthemetic progression with a=3 d=5
sum of n terms of an 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=5n²+n
=>5n²+n-2020=0
=>5n²+101n-100n-2020=0
=>n(5n+101)-20(5n+101)=0
=>(n-20)(5n+101)=0
=>n=20
I hope this will help u :)
the given series is in arthemetic progression with a=3 d=5
sum of n terms of an 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=5n²+n
=>5n²+n-2020=0
=>5n²+101n-100n-2020=0
=>n(5n+101)-20(5n+101)=0
=>(n-20)(5n+101)=0
=>n=20
I hope this will help u :)
dhathri123:
i hope u understand :)
Ohkie ohkie I got u thanks for ur help dear
welcome :)
Can u ans me one more question?
i will try
how many terms of the series 16+14+12+10+8+......must be taken so that their sum is zero
can u plzz ask the question
so i can answer
Yes I hv posted it pls ans me
ya i already answered it :)
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