Question

answer this question in attatchment

Answer 1

Answer:

77:176

Step-by-step explanation:

Let a₁,a₂ be the first terms and d₁,d₂ be the common differences of the two given AP's.

$=>\frac{\frac{n}{2}[2a_{1}+(n - 1) *d_{1}]}{\frac{n}{2}[2a_{2} + (n - 1) * d_{2}]} = \frac{3n + 8}{7n + 15}$

$=>\frac{2a_{1} + (n - 1) * d_{1}}{2a_{2} + (n - 1) * d_{2}} = \frac{3n+8}{7n+15}$

We need to find the ratio of their 12th term. So, put n = 23.

$=>\frac{2a_{1}+22d_{1}}{2a_{2}+22d_{2}} = \frac{3(23) + 8}{7(23) + 15}$

$=>\frac{a_{1}+11d_{1}}{a_{2}+11d_{2}} = \frac{77}{176}$

Therefore, ratio of their 12th terms is 77:176

Hope it helps!

Answer 2

▶ Question :-

The sum of first n term of two APs are in the ratio ( 3n + 8 ) : ( 7n + 15 ) . Find the ratio of their 12th terms .

▶ Answer :-

Let a , A be the first term and, d , D be the common difference of the two APs .

Suppose Sn and Sn' be the sum of the nth terms of the two APs .

°•° Sn = n/2[ 2a + ( n - 1 )d ] .

And Sn' = n/2[ 2A + ( n - 1 )D ] .

•°• Sn/Sn' = ( n/2[ 2a + ( n - 1 )d ] ) / ( n/2[ 2A + ( n - 1 )D ] ) .

= ( 2a + ( n - 1 )d ) / ( 2A + ( n - 1 )D ) .

==> ( 2a + ( n - 1 )d ) / ( 2A + ( n - 1 )D ) = ( 3n + 8 ) / ( 7n + 15 ) ....... [ Given ]

==> 2( a + ( n - 1 )/2 × d ) / 2( A + ( n - 1 )/2 × D ) = ( 3n + 8 ) / ( 7n + 15 ) .

==> ( a + ( n - 1 )/2 × d ) / ( A + ( n - 1 )/2 × D ) =
( 3n + 8 ) / ( 7n + 15 ).......(1) .

▶ Now, we need to find :- ( a + 11d )/( A + 11D ) .

Hence, ( n - 1 )/2 = 11 .

=> n - 1 = 22 .

•°• n = 23 .

Now, Putting the value of n = 23 in equation (1), we get

==> ( a + ( 23 - 1 )/2 × d ) / ( A + ( 23 - 1 )/2 × D ) =
( 3 × 23 + 8 ) / ( 7 × 23 + 15 ) .

==> ( a + ( 22/2 ) d ) / ( A + ( 22/2 ) D ) = ( 69 + 8 ) / ( 161 + 15 ) .

==> ( a + 11d ) / ( A + 11D ) = 77/176 .

✔✔ Hence, ratio of 12th term is 77 : 176 . ✅✅



$\boxed{ \boxed{ \pink{ \mathscr{ THANKS }}}}$