Question

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

Class X1 - Maths -Sequences and Series Page 185

Answer 1

Given,
Sn/S'n = (5n + 4)/(9n+6)

= n(5n+4)/n(9n+6)
=[n/2{10n +8}]/[n/2{18n+12}]
=[n/2{10n-10+10+8}]/[n/2{18n-18+18+12}]
=[n/2{10(n-1)+18}/[n/2{18(n-1)+30}]
=[n/2{2×9+10(n-1)}]/[n/2{2×15+18(n-1)}]

we know,
Sum of r terms = r/2{2a + (r-1)d}
Where, first term is a
And common difference is d .use it here,

Now,
for first AP series {Sn}:
first term (a) = 9
common difference (d) = 10

For second AP series {S'n}
First term (a') = 15
common difference (d') = 18

Now,

T18/T'18 = {a+(18-1)d}/{a'+(18-1)d'}
= {9 + 17×10}/{15+17×18}
= 179/321

Hence, ratio of their 18th terms = 179:321