Question

The sum of first n terms of two APs are in the ratio (3n+8):(7n+15). Find the ratio of their 12th terms.

Answer 1

Step-by-step explanation:

Let S1 be sum of first n terms of AP1 and S2 be the sum of first n terms of AP2

so

((n/2)*(2a1 +(n-1)d1))/((n/2)*(2a2 +(n-1)d2))=(3n+8)/(7n+15)

(2a1 +(n-1)d1)/(2a2 +(n-1)d2)=(3n+8)/(7n+15)

(a1 +(n-1)d1/2)/(a2 +(n-1)d2/2)...............................1

Since, T1n =a1 +(n-1)d1

Therefore,

from (1), (n-1)/2 =12

n-1 =24

n=25

so plug n=25 will give the required ratios of the nth terms of the two APs

(3*25+8)/(7*25+15)=(75+8)/(175+15)=83/190

Answer 2

Refers to attachment ♡~