Question

The ratio of the sum of n terms of two ap's(5n+3:3n+5) find the ratio of their nth term

Answer 1

Let there be two A.P.s:A.P.1 and A.P.2

Sn of A.P.1  : Sn of A.P.2=5n+3:3n+5

We know that Sn=n/2(2a+(n-1)d)

[n/2(2a+(n-1)d]:[n/2(2a+(n-1)d] = 5n+3: 3n+5            

=>(2a+(n-1)d : (2a+(n-1)d = 5n+3 : 3n+5

I we assume that the value of n is 2n-1

=>2a+(2n-1-1)d : 2a+(2n-1-1)d = 5(2n-1)+3 : 3(2n-1)+5

=>(2(a+(n-1)d)/(2(a+(n-1)d) = 5(2n-1)+3/3(2n-1)+5

=> a+(n-1)d/a+(n-1)d = 10n-2/6n-2=2(5n-1)/2(3n-1)

=> a+(n-1)d : a+(n-1)d = 5n-1 : 3n-1

Thus, nth term  of first A.P.1 : nth term of A.P. 2 = 5n-1 : 3n-1.

Thanks!!!

Answer 2

Answer:

Thus, nth term  of first A.P.1 : nth term of A.P. 2 = 5n-1 : 3n-1.

Step-by-step explanation:

Let there be two A.P.s:A.P.1 and A.P.2

Sn of A.P.1  : Sn of A.P.2=5n+3:3n+5

We know that Sn=n/2(2a+(n-1)d)

[n/2(2a+(n-1)d]:[n/2(2a+(n-1)d] = 5n+3: 3n+5            

=>(2a+(n-1)d : (2a+(n-1)d = 5n+3 : 3n+5

I we assume that the value of n is 2n-1

=>2a+(2n-1-1)d : 2a+(2n-1-1)d = 5(2n-1)+3 : 3(2n-1)+5

=>(2(a+(n-1)d)/(2(a+(n-1)d) = 5(2n-1)+3/3(2n-1)+5

=> a+(n-1)d/a+(n-1)d = 10n-2/6n-2=2(5n-1)/2(3n-1)

=> a+(n-1)d : a+(n-1)d = 5n-1 : 3n-1

Thus, nth term  of first A.P.1 : nth term of A.P. 2 = 5n-1 : 3n-1.