Question

how many terms of an ap 3 5 7 9 dotdotdot must be added to get the sum 120​

Answer 1

Answer:

Let the number of terms be 'n'.

A/q. n/2(2×3+(n-1)2)=120

n(3+n-1)=120

n^2+2n-120=0

n^2+12n-10n-120=0

n(n+12)-10(n+12)=0

(n+12)(n-10)=0

n+12 =0 or n-10=0

n=(-12.) or n=10

Required no of terms=10 terms

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