Question

if the sum of n terms of two AP are in the ratio (5n+3):(3n+4), then the ratio of their 17th term is

Answer 1

Given The sum of n terms of two AP are in the ratio (5n + 3) : (3n + 4).

To find : The ratio of their 17th term.

solution : we know, sum of n terms of an AP is given by, S = n/2[2a + (n - 1)d ]

so let's arrange in the standard form.

given, S₁/S₂ = (5n + 3)/(3n + 4)

= n(5n + 3)/n(3n + 4)

= [n/2(5n + 3)]/[n/2(3n + 4)]

= [n/2{5(n - 1) + 5 + 3}]/[n/2{3(n - 1) + 3 + 4}]

= [n/2{2 × 4 + (n - 1) × 5}]/[n/2{2 × 3.5 + (n - 1) × 3}]

so, for 1st ap, first term = 4 , common difference, d = 5

for 2nd ap, first term = 3.5 , common difference, d = 3

now nth term of 1st ap, T₁ = 4 + (n - 1)5

so, 17th term of 1st ap, T₁ = 4 + (17 - 1) × 5 = 4 + 80 = 84

again, nth term of 2nd ap, T₂ = 3.5 + (n - 1)3

so 17th term of 2nd ap, T₂ = 3.5 + (17 - 1)3 = 3.5 + 48 = 51.5

Therefore the ratio of their 17th term = T₁/T₂ = 84/51.5 = 168/103

Answer 2

Step-by-step explanation:

first temrs 4 second terms 5