Ratio of sum of two AP contains n terms is (3n+6):(5n-13).find their 11 terms
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Answer: Sum of two terms of an AP contains 'n' terms = (3n+6) : (5n-13)
11th term of an AP=?
= a+10d ÷ A+10D
Multiplying 2 on both sides
= 2a+20d ÷ 2A+20D
(n-1=20 , n=20+1=21 )
= sum of first 'n' terms =3n+6 ÷ sum of
sum of 2nd 'n' terms =5n-13
= n/2 (2a+(n-1)d=3n+6
n/2 (2a+(n-1)d=5n-13
= 2a+(21-1)d =69
2a+(21-1)d =92
= 2 (a+10d) =69
2 (A+10D)=92
= a+10d =69
A+10D=92
Therefore, 11th term of an ap 69:92.
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