Question

The sum of n terms of two A.P are in the ratio 5n-4:9n+6.find the ratio of their 18th terms

Answer 1

HELLO DEAR,



LET a1, a2, AND d1, d2 BE THE FIRST TERMS AND THE COMMON DIFFERENCE OF THE FIRST AND SECOND ARITHMETIC PROGRESSION RESPECTIVELY.

$\frac{s_{ n_{1} }}{ _{ s_{ n_{2}}} } = \frac{5n + 4}{9n + 6} \\ = > \frac{ \frac{n}{2}(2 a_{1} + (n - 1) d_{1} )}{ \frac{n}{2}(2 a_{2} + (n - 1) d_{2})} = \frac{(5n + 4)}{9n + 6)} .....................(1) \\ = > \frac{(a _{1} + 17 d_{1} ) }{ (a_{2} + 17 d_{2} )} = \frac{ a_{18}(first) }{ a_{18} (second) } \\ = > \frac{(2a _{1} + (35 - 1) d_{1} )}{(2a_{2} + (35 - 1) d_{2} )} = \frac{5 \times 35 + 4}{9 \times 35 + 6}.....using(1) \\ = > \frac{a_{18}(first)}{a_{18} (second)} = \frac{179}{321}$
THUS, THE RATIO OF 18th TERM OF BOTH The A.P.s is 179: 321.


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