google is your friend:
http://m.nextgurukul.in/nganswers/ask-question/answer/The-ratio-of-the-sum-of-n-terms-of-two-APs-is-7n14n27/Arithmetic-Progressions/21214.htm
The sum of n tearm of two arithmetic progressions are in the ratio (7n+1):(4n+27).Find the ratio of their 11th terms
2 answers
for 1st series:
first term is a, common difference is d
sum(n) = (n/2)(2a + d(n-1))
for 2nd series:
first term is b, common difference is e
sum(n) = (n/2)(2b + e(n-1))
(n/2)(2a + d(n-1)) / (n/2)(2b + e(n-1)) = (7n+1)/(4n+27)
(2a + dn - d)/ (2b + en- e) = (7n+1)/(4n+27)
8an + 54a + 4dn^2 + 27dn - 4nd - 27d = 14bn + 7en^2 + 2b + en - e ***
You also know that n ≥ 11, and n must be a whole number
ratio of 11 terms = (a+10d)/(b+10e) **
Does that give you something to play with ?
first term is a, common difference is d
sum(n) = (n/2)(2a + d(n-1))
for 2nd series:
first term is b, common difference is e
sum(n) = (n/2)(2b + e(n-1))
(n/2)(2a + d(n-1)) / (n/2)(2b + e(n-1)) = (7n+1)/(4n+27)
(2a + dn - d)/ (2b + en- e) = (7n+1)/(4n+27)
8an + 54a + 4dn^2 + 27dn - 4nd - 27d = 14bn + 7en^2 + 2b + en - e ***
You also know that n ≥ 11, and n must be a whole number
ratio of 11 terms = (a+10d)/(b+10e) **
Does that give you something to play with ?
1 answer