Question

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Answer 1

➛G I V E N:-

•First term of AP=1

•Common difference of the AP=4

•Sum of the nth term of AP=190

➛T O F I N D O U T :-

•The number of terms in that AP(n)=0

➛S O L U T I O N :-

•Let a= first term

d=common difference

Sn=Sum of nth term of AP

Now we have a=1 ,d=4 and Sn=190

We know that

•Sn=n/2[2a+(n-1)d]

⇢190=n/2[2×1+(n-1)4]

⇢2×190=n[2+4n-4]

⇢380=n[4n-2]

⇢380=4n²-2n

⇢4n²-2n-380=0

⇢2n²-n-190=0

Factorising it by mid term splitting method

⇢2n²-20n+19n-190=0

⇢2n(n-10)+19(n-10)=0

⇢(2n+10)(n-10)=0

⇢2n+10=0 or n-10=0

⇢n=-10/2=-5 or n=10

•But the terms in AP cannot be negative , neglecting -5 , we have n=10

Therefore, there are 10 th terms in the given A•P