Latest TCS Aptitude Question SOLUTION: 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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(#M40012839) TCS Logical Reasoning question Keep an EYE

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

Asked In TCS (11 years ago)

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Read Solution (8) :

raj  (11 years ago)

Ratio of sum of n terms 2 A.P = 5n+4 : 9n+6

So, Ratio Sum of 13 terms = 5*13+4 : 9*13+4 (put n=13)
= 69 : 123
Now, Ratio Sum of 12 terms = 5*12+4 : 9*12+4 (put n=12)
= 64 : 114

So to claculate 13th term we will simple subtract sum of first 12 terms from sum of first 13 terms

i.e, Sum(13) - Sum(12) = = 69 : 123 - 64 : 114

(The important thing is tht we will not subtract it by taking LCM as these are ratios, We will subtract Numerator to Numerator & Denomenator to Denomenator)
So,
Ratio of 13th term of 2 APs will be = 5 : 9

Is this solution Helpfull? Yes (47) | No (21)  

Ajit Ankit Asheesh Anurag  (11 years ago)

given, 2A+(n-1)D:2a+(n-1)d= 5n+4:9n+6
now, put n-1=24...i.e. n=25
we get, 2A+24D:2a+24d= 2*(13th term of 1st ap):2*(13th term of 2nd ap)
hence ratio= (5*25+4):(9*25+6)=129:231

Is this solution Helpfull? Yes (13) | No (2)  

Sarvesh Shukla  (11 years ago)

Ans-5:9
Ratio of sum of n terms of 2 A.P = 5n+4:9n+6 (given)
now, ratio of sum of 13terms S(13)= 69:123 (put n=13)
and ratio of sum of 12 terms S(12)= 64:114 (put n=12)
So, 13th term [S(13)-S(12)]= 5:9

Is this solution Helpfull? Yes (7) | No (5)  

Rahul Raj  (11 years ago)

129/231
Sn=n*middle term
let 13th term is middle term then total no. of terms =25.put n=25

Is this solution Helpfull? Yes (6) | No (10)  

vigneshwari  (11 years ago)

Ajit Ankit, pls explain
how it came n-1 = 24

Is this solution Helpfull? Yes (4) | No (2)  

vigneshwari  (11 years ago)

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 solution Helpfull? Yes (4) | No (3)  

Priya  (8 years ago)

We should bring a solution that will be

13th term of 1st AP : 13th term of 2nd AP

=> a+(13-1)d of AP 1 / a+(13-1)d of AP 2

=> 2[ a+(13-1)d] of AP 1 / 2[a+(13-1)d] of AP 2
=> 2a+(12*2)d of AP 1 / 2a+(12*2)d of AP 2
=> 2a + 24d of AP 1 / 2a+ 24d of AP 2
=> n=25 because for n=25, sum of n terms is 2a+ 24d

thus substitute n=25
then,
5*25+4 / 9*25+6
=>129/231.

Is this solution Helpfull? Yes (2) | No   

karthikeyank  (9 years ago)

Ans-5:9
Ratio of sum of n terms of 2 A.P = 5n+4:9n+6 (given)
now, ratio of sum of 13terms S(13)= 69:123 (put n=13)
and ratio of sum of 12 terms S(12)= 64:114 (put n=12)
So, 13th term [S(13)-S(12)]= 5:9

Is this solution Helpfull? Yes (1) | No (1)  

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