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