Question

If ratio of sum of 1 at n term of 2 different AP= 2n-3/3n+1 find the ratio of the 9 th term of those 2A.P

Answer 1

Answer = - 23/4

Step-by-step explanation:

Let the first AP be {a, a+d, a+2d,........upto n terms}

Let the 2nd AP be {b, b+D, b+2D,.......... Upto n terms}

Ratio of their sum of n terms of two AP =2n-3/3n+1

Let the ratio be in form of some constant k, then ratio=(2n-3)k/(3n+1)k

Now put n=1 in the ratio

We get Sⁿ {ratio of sum of AP} = - k/4k —(i)

Now on putting n=2

Sⁿ = 1k/7k

According to above equation....

a+a+d/b+b+D = 1k/7k

On solving and comparing

2a+d =k

d=k-2a, on further solving d=k-(-2k).

d=3k

Required AP is {-k, 2k, 5k.............}

9th term of this AP =a+8d=-k+24k=23k

Now for taking out series of b

b=4k {from eq(I)}

2b+D=7k(solved above and now on comparing)

D=7k-2b, =-k

Now the req AP is {4k, 3k, 2k,........}

9 th term of this AP =b +8D,

=4k-8k=-4k

Ratio of their 9th term = 23k/-4k, or 23/4