The sum of n terms of two Arithmetic Progressions are in the ratio 7n+1 : (4n 17)

CBSE 10 - Maths

Asked by aayeshashaikh40.10 | 24 Jul, 2022, 06:42: PM

Text Solution

Solution : Given, `S_(n):S'_(n)=(7n+1):(4n+17)` <br> Here, `A=7,B=1,C=4 and D = 17 ` <br> `therefore (t_(n))/(t'_n)=(7(2n-1)+1)/(4(2n-1)+17)=(14n-6)/(8n+13)` <br> and `(d)/(d')=(A)/(C)=(7)/(4)` <br> Hence, `t_(n):t'_(n)=(14n-6):(8n+13)` and `d:d'= 7:4`

Text Solution

Solution : Given, `S_(n):S'_(n)=(7n+1):(4n+17)` <br> Here, `A=7,B=1,C=4 and D = 17 ` <br> `therefore (t_(n))/(t'_n)=(7(2n-1)+1)/(4(2n-1)+17)=(14n-6)/(8n+13)` <br> and `(d)/(d')=(A)/(C)=(7)/(4)` <br> Hence, `t_(n):t'_(n)=(14n-6):(8n+13)` and `d:d'= 7:4`