Question

Q.2) The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms.

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Answer 1

Answer:

Ratio of their 18th terms are 179/321

Step-by-step explanation:

Sum of n terms of 1st ap / Sum of n terms of 2nd AP

= 5n+4/9n+6

Let 1st term of 1st ap be a and 1st term of 2nd ap be A.

n/2 [2a+(n-1)d] / n/2[2A+(n-1)d] = 5n+4/9n+6

2[a+(n-1)/2 d] / 2[A+(n-1)/2 d] = 5n+4/9n+6

We are asked to find ratio of 18th terms ,they are

a+17d and A+17d

So in above equation let's substitute n as 35

Because (n-1)/2 = 17

(n-1)=34

n=35

[a+17d] / [A+17d] = 5(35)+4/9(35)+6

= 175+4/315+6

=179/321

Therefore, required ratio is 179/321

Answer 2

$\huge\tt\colorbox{aqua}{Solution:}$

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

$\sf \pink{ \underline{For \: the \: second \: AP \: , the \: first \: term \: is \: A \: and \: the \: common \: difference \: is \: D,}}$

$\sf \pink{ \underline{Now \: as \: per \: the \: problem \: ,}}$

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

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

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

$\sf \pink{ \underline{Now \: the \: ratio \: of \: 18th \: term \: of \: both \: the \: ap \: is}}$

$=\frac{a+17 d}{A+17 D}$….. (ii)

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

$\sf \pink{ \underline{n=35}}$

$\sf \pink{ \underline{Now \: putting \: value \: of \: n \: in \: equation \: (i), \: we \: get}}$

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

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

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

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