Question

If the ratio of the sum of first of two A.P.'s is (7n+1):(4n+27), find the ratio of their m th terms​

Answer 1

Answer:

The ratio of two AP's mth terms = 14m - 6 : 8m + 23 .

Step-by-step explanation:

We are given that the ratio of the sum of the first n terms of two A.P's is (7n + 1) : (4n + 27) i.e.,

                                $\frac{Sum of first n terms of 1st A.P.}{Sum of first n terms of 2nd A.P.} = \frac{7n+1}{4n+27}$

We know that sum of first n terms of an AP = $\frac{n}{2}[2a + (n-1)d]$ .

So, here let first term of first AP = $a_1$ and first term of second AP = $a_2$ and also common difference of first AP = $d_1$ and common difference of second  AP = $d_2$ .

              ⇒  $\frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{7n+1}{4n+27}$

              ⇒  $\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{7n+1}{4n+27}$

              ⇒  $\frac{a_1 + (\frac{n-1}{2})d_1 }{a_2 + (\frac{n-1}{2})d_2 } = \frac{7n+1}{4n+27}$  -------------- [Equation 1]

Now, we have to find the ratio of two AP's mth terms which is given by;

                             $\frac{a_1 + (m-1)d_1}{a_2 + (m-1)d_2}$

Now, putting $\frac{n-1}{2} = m-1$  we get, n - 1 = 2*m - 2 ⇒ n = 2*m - 1 .

To make this ratio comparable with equation 1 , put n = 2*m - 1 in RHS ;

So,        $\frac{a_1 + (m-1)d_1}{a_2 + (m-1)d_2} = \frac{7*(2m-1)+1}{4*(2m-1)+27} = \frac{14m -7+1}{8m-4+27}$       

                               = $\frac{14m -6}{8m+23}$

Hence, the ratio of two AP's mth terms = 14m - 6 : 8m + 23 .