Question

the sum of n terms of two arithmetic progression are in the ratio 7n + 1 is to 4n+ 27. find the ratio of the 11th terms.​

Answer 1

Answer:

⇒ 148 : 111

Step-by-step explanation:

Given :

The sum of n terms of two arithmetic progression are in the ratio 7n + 1 is to 4n+ 27.

Find the ratio of the 11th terms.​

Solution :

We know that,

Sum of n terms of an AP is given by,

⇒ $S_n=\frac{n}{2}(2a+(n-1)d)$

Let the first term , common difference & sum upto n terms of first AP be $A_1$ , $D_1$ & $S_{n}$ respectively,.

Let the first term & common difference & sum upto n terms of 2nd AP be $A_2$ , $D_2$  & $S'_{n}$ respectively,.

then,

We know that,.

$\frac{S_{n1}}{S_{n2}} = \frac{7n+1}{4n+27}$

So,.

We also knw that,.

nth term of an AP is given by :

$a_n=a+(n-1)d$

So, their ratio in nth terms is :

⇒ $\frac{A_{n1}}{A_{n2}} = \frac{A_1+(n-1)D_1}{A_2+(n-1)D_2}$

By multipying & dividing by  in RHS,.

⇒ $\frac{A_{n1}}{A_{n2}} = \frac{2A_1+2(n-1)D_1}{2A_2+2(n-1)D_2}$

⇒ $\frac{2A_1+(2n-2)D_1}{2A_2+2(n-1)D_2}$

⇒ $\frac{2A_1+[(2n-1)-1]D_1}{2A_2+[(2n-1)-1]D_2}$

By multiplying by $\frac{2n-1}{2}$ on both numerator & denominator on RHS,.

⇒ $\frac{\frac{2n-1}{2}(2A_1+[(2n-1)-1]D_1)}{\frac{2n-1}{2}(2A_2+[(2n-1)-1]D_2)}$

⇒ $\frac{S_{2n-1}}{S'_{2n-1}}$

⇒ $\frac{7(2n-1)+1}{4(2n-1)+27} = \frac{14n-7+1}{8n-4+27} = \frac{14n-6}{8n+23}$

Thus the ratio of nth terms of two AP’s is [14n – 6] : [8n + 23].

Substituting n = 11,

We get,

⇒ [14(11) – 6] : [8(11) + 23]

⇒ [154 – 6] : [88 + 23]

⇒ 148 : 111