The ratio of the sum of n terms of two A.P.'s be `5(2n+1):(97-2n)`. Find (i) the ratio of their nth terms. Hence, find the ratio of their 16th terms. (ii) If first term and common difference of first A.P. are 3 and 4 respectively, find the second A.P.

Explanation

(i) Let first term and common difference of one A.P. be `a_(1) and d_(1)` and let first term and common difference of second A.P. be `a_(2) and d_(2)`.
`:. (S_(n))/(S_(n)^(**))=(5(2n+1))/(97-2n) " " `(given)
`rArr ((n)/(2)[2a_(1)+(n-1)d_(1)])/((n)/(2)[2a_(2)+(n-1)d_(2)])=(5(2n+1))/(97-2n) " " ` (using formula for sum of n terms)
`rArr (2a_(1)+(n-1)d_(1))/(2a_(2)+(n-1)d_(2))=(5(2n+1))/(97-2n)`
`rArr (a_(1)+((n-1)/(2))d_(1))/(a_(2)+((n-1)/(2))d_(2))=(5(2n+1))/(97-2n)` ...(1) (dividing each term of Nr. and Dr. by 2)
Now, we want the ratio of nth terms i.e., `(T_(n))/(T_(n)^(**))=(a_(1)+(n-1)d_(1))/(a_(2)+(n-1)d_(2))`
So, replace `(n-1)/(2) rarr n-1`
`rArr n-1 rarr 2n-2`
`rArr n rarr 2n-2+1`
`rArr n rarr 2n-1`
`:. ` Eq. (1) becomes,
`(a_(1)+((2n-1-1)/(2))d_(1))/(a_(2)+((2n-1-1)/(2))d_(2))=(5[2(2n-1)+1])/(97-2(2n-1))`
`:. " " (a_(1)+(n-1)d_(1))/(a_(2)+(n-1)d_(2))=(5(4n-1))/(99-4n)`
`rArr (T_(n))/(T_(n)^(**))=(5(4n-1))/(99-4n) " " ...(2)`
Hence, the ratio of nth terms is 5 (4n-1) : (99-4n).
Put n=16 in the above relation (2), we get
`:. (T_(16))/(T_(16)^(**))=(5(4xx16-1))/(99-4xx16)=(5xx63)/(35)=9:1`
(ii) `a_(1)=3, d=4`
`:.` First A.P. is` 3,3+4,3+2(4),3+3(4),...`
i.e., 3, 7, 11, 15, 19, 23, ...,
Put n=1 in eq. (2),
`(T_(1))/(T_(1)^(**))=(5(4-1))/(99-4)=(5xx3)/(95)=(3)/(19)`
`rArr (3)/(T_(1)^(**))=(3)/(19) rArr T_(1)^(**)=19`
Arithmetic Progression
Put n=2 in eq. (2),
`rArr (T_(2))/(T_(2)^(**))=(5(8-1))/(99-8)=(5xx7)/(91)=(5)/(13)`
`rArr (7)/(T_(2)^(**))=(5)/(13) rArr T_(2)^(**)=(91)/(5)=18(1)/(5)`
`:. d-T_(2)^(**)-T_(1)^(**)=(91)/(5)-19=-(4)/(5)`
`:.` Second A.P. is `T_(1)^(**), T_(2)^(**), T_(3)^(**), ...`
i.e., `19, 18(1)/(5),17(2)/(5),16(3)/(5),... " " ` `((T_(3)^(**)=A_(2)+d),(^(**)T_(4)^(**)=a_(3)+d^(**).....))`