The-sums-of-the-first-n-terms-of-two-AP-s-are-in-the-ratio-3n-31-5n-3-Show-that-their-9-th-terms-are-equal-

Question Number 2619 by Rasheed Soomro last updated on 23/Nov/15

T h e s u m s o f t h e f i r s t n t e r m s o f t w o A P^{'} s a r e

i n t h e r a t i o 3 n + 31 : 5 n - 3 . S h o w t h a t t h e i r 9^{t h} t e r m s

a r e e q u a l .

Commented by Yozzi last updated on 24/Nov/15

S_{1} (n) = \frac{n}{2} (a_{1} + l_{1}) = \frac{{n c}_{1}}{2}, S_{2} (n) = \frac{n}{2} (a_{2} + l_{2}) = \frac{{n c}_{2}}{2}

\frac{S_{1} (n)}{S_{2} (n)} = \frac{3 n + 31}{5 n - 3} = \frac{c_{1}}{c_{2}} .

3 c_{2} n + 31 c_{2} = 5 c_{1} n - 3 c_{1}

n (3 c_{2} - 5 c_{1}) = - 3 c_{1} - 31 c_{2}

n = \frac{3 c_{1} + 31 c_{2}}{5 c_{1} - 3 c_{2}}

n - 1 = \frac{34 c_{2} - 2 c_{1}}{5 c_{1} - 3 c_{2}}

Δ = T_{1} (9) - T_{2} (9) = a_{1} - a_{2} + 8 (d_{1} - d_{2})

l_{1} = a_{1} + (n - 1) d_{1} \Rightarrow d_{1} = \frac{l_{1} - a_{1}}{n - 1}

δ = d_{1} - d_{2} = \frac{l_{1} - a_{1} - l_{2} + a_{2}}{n - 1}

δ = \frac{(l_{1} + a_{2} - a_{1} - l_{2}) (5 c_{1} - 3 c_{2})}{34 c_{2} - 2 c_{1}}

Answered by prakash jain last updated on 24/Nov/15

Let 2 a_{1} + (n - 1) d_{1} = (3 n + 31) k_{1}, then

2 a_{2} + (n - 1) d_{2} = (5 n - 3) k_{1}

(2 a_{1} - d_{1}) + {n d}_{1} = 3 {n k}_{1} + 31 k_{1} \dots (1)

If (1) is true for all n (c o e f f i i e n t m u s t b e e q u a l)

{n d}_{1} = 3 {n k}_{1} \Rightarrow d_{1} = 3 k_{1}

2 a_{1} - d_{1} = 31 k_{1} \Rightarrow a_{1} = 17 k_{1}

similarly

d_{2} = 5 k_{1}

a_{2} = k_{1}

9 t h term

a_{1} + 8 d_{1} = 17 k_{1} + 24 k_{1} = 41 k_{1}

a_{2} + 8 d_{2} = k_{1} + 40 k_{1} = 41 k_{1}

So 9 t h terms are equal .

Answered by Rasheed Soomro last updated on 24/Nov/15

W i t h o u t r e p e a t i n g u n d e r s t o o d t h i n g s

\frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{3 n + 31}{5 n - 3} [G i v e n]

a_{1} + 8 d_{1} (T_{9}) = a_{2} + 8 d_{2} (t_{9}) [R e q u i r e d]

\frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{3 n + 31}{5 n - 3}

O u r g o a l : T o a c h i e v e a_{1} + 8 d_{1} i n n u m e r a t o r

a n d a_{2} + 8 d_{2} i n d e n o m i n a t o r b e c a u s e w e w a n t

t o d e t e r m i n e r e l a t i o n b e t w e e n T_{9} a n d t_{9} .

\Rightarrow \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{3 n + 31}{5 n - 3}

\Rightarrow \frac{2 [a_{1} + (\frac{n - 1}{2}) d_{1}]}{2 [a_{2} + (\frac{n - 1}{2}) d_{2}]} = \frac{3 n + 31}{5 n - 3}

\Rightarrow \frac{a_{1} + (\frac{n - 1}{2}) d_{1}}{a_{2} + (\frac{n - 1}{2}) d_{2}} = \frac{3 n + 31}{5 n - 3}

W e w a n t 8 i n p l a c e o f \frac{n - 1}{2}

∴ \frac{n - 1}{2} = 8 \Rightarrow n = 17

\Rightarrow \frac{a_{1} + 8 d_{1}}{a_{2} + 8 d_{2}} = \frac{3 (17) + 31}{5 (17) - 3} = \frac{82}{82} = 1

∴ T_{9} = t_{9}

Commented by prakash jain last updated on 24/Nov/15

This method is simpler .