Question

The sums of n term of AP are in the ratio (7n-5) :(5n+17)show that their 6th term are equal

Answer 1

Answer:

The sums of n terms of two arithmetic progressions are in the ratio (7n-5) : ( 5n+17) . Show that their 6th terms are equal. Hence,their 6th terms are equal.

Answer 2

Answer:

here is ur answer dear

Let a , d and A , D be the first term and the common ratio of the first and the second AP respectively.

So [(n/2){2a+(n-1)d}]/[(n/2){2A+(n-1)D}]

=(7n-5)/(5n+17)={7(n-1)+2}/{5(n-1)+22}

So {2a+(n-1)d}/{2A+(n-1)D}

={2+7(n-1)}/{22+5(n-1)} =>

a=1, d=7, A =11, D = 5

6th term of 1st AP =a+5d=1+5×7=36

6th term of 2nd AP=A+5D=11+5×5=36

hope it helps u dear...I hope so that this is correct and helps u dear...

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