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If the sum of n terms of two series of A.P are in the ratio
5n+4:9n+6 .find the ratio of their 13th terms
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Answer / krishna kumar ranjan
when ever u face this type of qun..
13*2-1=25 and put it in place of n in 5n+4/9n+6
| Is This Answer Correct ? | 61 Yes | 30 No |
Answer / akshay b menon
(5n+4)/(9n+6)=((n/2){2x+(n-1)d})/((n/2){2x'+(n-1)d'})
=(2x+(n-1)d)/(2x'+(n-1)d')
=(x+{(n-1)/2}d)/(x'+{(n-1)/2}d')
implies,
m=(n-1)/2
2m-1=n
since m=13,n=25
substituting value of n in given ratio
(5*25+4)/(9*25+6)=129/231
| Is This Answer Correct ? | 45 Yes | 47 No |
Answer / narayanan
Let the 2 series have the sum S(1,n) and S(2,n).
now, S(1,n):S(2,n) = (5n+4):(9n+6)
= {5(n-1)+5+4}:{9(n-1)+9+6}
= {2*(9/2)+5(n-1)}:{2*(15/2)+9(n-1)}
= [(n/2){2*(9/2)+5(n-1)}]:[(n/2){2*(15/2)+9(n-1)}]
hence nth terms will be in the ratio
T(1,n):T(2,n)= [(9/2)+5(n-1)]:[(15/2)+9(n-1)]
=>T(1,13):T(2,13)=[(9/2)+5(13-1)]:[(15/2)+9(13-1)]
=(129/2):(231/2)
=43/77
| Is This Answer Correct ? | 50 Yes | 67 No |
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