Question

the ratio between the sum of n terms of two aps is (3n+1)(4n+5)the ratio between the 10th term is​

Answer 1

this question is tur sorry

Answer 2

• sum of nth term of an ap can be written as - n/2(2a+(n-1)×d) here a is first term and d is common diffrence
• nth term =a+(n-1)×d.
• 10 th term is written as -

a+9d or a+(a+8d).

let first term of both AP are a1 and a2

and common diffrence is d1 & d2

[n/2(2a1+(n-1)d1)] /[n/2(2a2+(n-1)d2] =(3n+1)/(4n+5)

after cancelling the n/2 we get

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

=(3n+1)/(4n+1)

it can be written as- a1+9d1 for 10 th term and amsame for second AP

so value of n=9 and substituting n=9 on right hand side we get 28/41 which is required answer