Question

The sums of n terms of two arithmetic progressions are in the ratio 5n+4: 9n+6. Find the ratio of their 18th terms.

Answer 1

Answer:

The ratio of $18^{\text {th }}$ term of both the AP is 179:321

Solution:

Let us assume that for the first AP, the first term is a and common difference is d

For the second AP, the first term is A and the common difference is D,

Now as per the problem,

$\frac{\frac{n}{2}[2 a+(n-1) d]}{\frac{n}{2}[2 A+(n-1) D]}=\frac{5 n+4}{9 n+6}$

$\frac{[2 a+(n-1) d]} {[2 A+(n-1) D]}=\frac{5 n+4}{9 n+6}$

$\frac{\left[a+\frac{(n-1) d}{2}\right]}{\left[A+\frac{(n-1) D}{2}\right]} \: =\frac{5 n+4}{9 n+6} ……………….. (i)$

Now the ratio of 18th term of both the ap

is :-

$=\frac{a+17 d}{A+17 D}….. (ii)$

Hence ,

$\frac{n-1}{2}=17$

$n=35$

Now putting value of n in equation (i), we get

$\frac{\left[a+\frac{(35-1) d}{2}\right]}{\left[A+\frac{(35-1) D}{2}\right]}=\frac{5 \times 35+4}{9 \times 35+6}$

$\frac{a+17 d}{A+17 D}=\frac{179}{321}$

So the ratio of $18^{\text {th }}$term of both the AP is 179:321

Hope it helps you

@itzheartcracer

Answer 2

Answer:

The ratio of their 18th  terms is $179:321$.

Step-by-step explanation:

Given :-

Sum of n terms of two arithmetic progressions are in the ratio 5n + 4, 9n + 6.

To find :-

Ratio of their 18th terms.

Solution :-

$S=\frac{n}{2}[2a+(n-1)d]$ , $\frac{n}{2}[2A+(n-1)D]$.

$\rightarrow \frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2A+(n-1)D]}=\frac{5n+4}{9n+6}$

$\rightarrow \frac{2a+(n-1)d}{2A(n-1)D}=\frac{5n+4}{9n+6}$

$a_1_8=a+17d$ , $A_1_8=A+17D$. (18 + 17 = 35), $n=35$.

$\rightarrow\frac{2a+34d}{2A+34D}=\frac{5(35)+4}{9(35)+6}$

$\rightarrow\frac {2[a+17d]}{2[A+17D]}=\frac{179}{321}$

$\rightarrow\frac{a+17d}{A+17D}=\frac{179}{321}$

$\rightarrow\frac{a_1_8}{A_1_8}=\frac{179}{321}$

∴ The ratio of their 18th  terms is $179:321$.