The sum of n terms of two arithmetic progressions are in the ratio 5n+4: 9n+6. Find the ratio of their 18th terms.

Answer

N/a

Explanation

Let the first terms be `a_(1)" and "a_(2)` and common difference be `d_(1)" and "d_(2)` respectively of two A.P.'s.
Now, according to the problem,
`((n)/(2)[2a_(1)+(n-1)d_(1)])/((n)/(2)[2a_(2)+(n-1)d_(2)])=(5n+4)/(9n+6)`
`rArr" "(2a_(1)+(n-1)d_(1))/(2a_(2)+(n-1)d_(2))=(5n+4)/(9n+6)" ...(1)"`
Now, ratio of 18th terms of two progressions
`(t_(18))/(T_(18))=(a_(1)+17d_(1))/(a_(2)+17d_(2))=(2a_(1)+34d_(1))/(2a_(2)+34d_(2))`
`=(2a_(1)+(35-1)d_(1))/(2a_(2)+(35-1)d_(2))=(5xx35+4)/(9xx35+6)`
`(179)/(321)" [put n = 35 in equation (1)]"`
Therefore, ratio of 18th terms of two progressions is 179 : 321.