Question

Two A.P are given 9,7,5,.......and 24,21,18,.....

If n th term of both the progression are equal,

then find the value of n and n th term.​

Answer 1

Solution:-

Given:-

Two APs are given 9, 7, 5... and 24, 21, 18....

and nth term of both the progression are equal

Find:-

nth term = ?

value of n = ?

________________________

Let the first term of 1st AP be a and the first term of 2nd AP be b...

Also let the common difference of first AP be d1 and for 2nd AP be d2...

So, by following this...

For 1st AP that is 9, 7, 5....

first term(a)= 9

common difference(d1)= 7 - 9 = -2

So, nth term for the first AP is given by...

$\sf{a_n = a + (n - 1)d_1}$.......eq(1)

For 2nd AP that is 24, 21, 18.....

first term(b) = 24

common difference(d2) = 21 - 24 = -3

nth term of second AP is given by...

$\sf{a_n = b + (n - 1)d_2}$.....eq(2)

So, by following this

from equation 1 and 2...

we will get,

$\sf{a + (n - 1)d_1 = b + (n - 1)d_2}$

$\sf{9 + (n - 1)(-2) = 24 + (n - 1)(-3)}$

$\sf{9 - 2n + 2 = 24 - 3n + 3}$

$\sf{11 - 2n = 27 - 3n}$

$\sf{11 - 27 = -3n + 2n}$

$\sf{-16 = -n}$

n = 16

by putting this in equation (1)

a(n) = a + (n - 1)d1

a(n) = 9 + (16 - 1)(-2)

a(n) = 9 + (15)(-2)

a(n) = 9 - 30

a(n) = -21

_______________________________

Answer 2

Answer:

let tn is the n th term of both the A.P

For the A.P 9,7,5

t1=9

d=7-9=-2

tn = a1+(n-1)d

tn =9+(n-1)×-2

tn = 9-2n+2

tn = 11-2n............(1)

And for A.P

24,21,18

a2=24

d2=21-24=-3

tn = a2+(n-1)d

tn = 24+(n-1)×-3

tn = 24-3n +3

tn = 27-3n.....(2)

from 1 &2

11-2n=27-3n

-2n+3n = 27 -11

n=16

using the eq.(1)

we get,

tn = 11-2×16

tn =11-32

tn = -21.