Question

If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27),

then find the ratio of their 9th terms.​

Answer 1

Answer:

The ratio of two AP's 9th terms = 24 : 19.

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 = $d_2$ .

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

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

Now, we have to find the ratio of two AP's 9th terms which is given by;

             $\frac{a_1 + (n-1)d_1}{a_2 + (n-1)d_2}$ = $\frac{a_1 + (9-1)d_1}{a_2 + (9-1)d_2}$ = $\frac{a_1 + 8d_1}{a_2 + 8d_2}$     {As $a_n = a + (n-1)d$}

To make this ratio comparable with equation 1 , multiply numerator and denominator of this ratio by 2 i.e.,

         ⇒ $\frac{2a_1 + 16d_1}{2a_2 + 16d_2} = \frac{2a_1 + (17-1)d_1}{2a_2 + (17-1)d_2}$

Now, After comparing the above ratio with Equation 1 we observe that the value of n comes out to be 17.

So,  $\frac{a_1 + 8d_1}{a_2 + 8d_2}$ = $\frac{7*17+1}{4*17+27}$ = $\frac{120}{95}$

Hence, the ratio of two AP's 9th terms = 24 : 19 .