CBSE - X - Mathematics - Arithmetic Progression
About an A.P sum.............
Your question was
The ratio of the sum of n terms of two AP is (4n+2): (3n+47).find the ratio of the 9th term.
Let the I term and common difference of the AP be a and d and for the II AP it will be
a' and d' respectively
According to the given Condition
n/2(2a+(n-1)d): n/2(2a'+(n-1)d') = (4n+2): (3n+47)
i.e (2a+(n-1)d): (2a'+(n-1)d') =(4n+2): (3n+47)
i.e (a+(n-1)d/2):(a'+(n-1)d/2)=(4n+2): (3n+47)
substituting n=17 in both sides we get
(a+8d):(a'+8d') = 70:98
(a+8d):(a'+8d') = 5:7
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