Question

the ratio of sum of n terms for two ap's (5n +4):(9n+6) find the their 18th term​

Answer 1

Answer:

Step-by-step explanation:

Given:

ratio of sum of n terms of an A.P =$\dfrac{(5n+4)}{(9n+6)}$

pre-requisite knowledge

formula for sum of 'n' terms =$\dfrac{n}{2}(2a+(n-1)d)$

formula for 'nth term=$a+(n-1)d =a_n$

solution:

let 'a' be the first term 'd' be the common difference for the first a.p and for the second a.p let it be 'p' and 'q' respectively$\therefore\:according \:to\:the\:question\:\\\dfrac{s_n}{s_n}=\dfrac{2a+(n-1)d}{2p+(n-1)q}=\dfrac{5n+4}{9n+6}\\\\let's\:divide\:the\:L.H.S\:by\:2\\\implies\dfrac{a+\dfrac{n-1}{2}d}{p+\dfrac{n-1}{2}q}=\dfrac{5n+4}{9n+6}\\we\:have\:to\:find\:18th\:term\:ratio\\\implies\dfrac{a+(18-1)d}{p+(18-1)d}=\dfrac{a+17d}{p+17d}=we\:have\:to\:find\\Hence\\\dfrac{n-1}{2}=17\\\implies\:n=35\\puting\:the\:value\:in\:first\\\dfrac{a+\dfrac{35-1}{2}d}{p+\dfrac{35-1}{2}q}=\dfrac{5*35+4}{9*35+6}\\\implies\:\dfrac{179}{321}$

if 'x' is the common factor then 18 term 179x and321x