Question

# The sum of n terms of two A.P. are in the ratio 5n+4:9n+6. The ratio of their 18th terms is

A
47:84
B
179:321
C
89:159
D
2:3

Solution

## The correct option is B 179:321Let the sum of nth terms of two AP be denoted by sn and Sn. The first term and common difference of two AP's be denoted by a & d and A & D respectively snSn=5n+49n+6⇒n2(2a+(n−1)d)n2(2A+(n−1)D)=5n+49n+6⇒(2a+(n−1)d)(2A+(n−1)D)=5n+49n+6Dividing Numerator and denominator by 2,a+(n−1)2dA+(n−1)2D=5n+49n+6 The ratio of 18th term of two A.P. is a18A18=a+17dA+17DComparing with a+(n−1)2dA+(n−1)2D=5n+49n+6 ...(1] we get,(n−1)2=17⇒n=35Putting n=35 in (1], we geta18A18=5(35)+49(35)+6=179321

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