Question

In an AP Tn+5
=35 and Tn+1
=23 then the common difference and first term is?

Answer 1

Tn+5=35
a+(n+5-1)d=35
a+nd+4d=35
Tn+1=23
a+(n+1-1)d=23
a+nd=23
substrate eq 1 and 2
a+nd+4d-a-nd=35-23
4d=12
d=3

Answer 2

Given,

Tn+ 5 = 35

Tn+1 =23

To Find,

d= ?

a =?

Solution,

from the formula of nth term in AP,

$a_n = a +(n-1)d$

$a_{n+5} = a +(n + 5 -1)d = 35$

$a +(n + 4)d = 35$ [Equation 1 ]

Similarly,

$a_{n+1} = a +(n + 1 -1)d = 23$

$a_{n+1} = a +(n )d = 23$[Equation 2]

Subtracting equation 1 from equation 2

$a + (n + 4)d - (a + nd) = 35 - 23$

$a + nd + 4d - a - nd = 12$

$4d = 12\\d =12/4\\d = 3$

Common difference(d) = 3

For nth term,

$a_n = a + (n-1)d = a + 3d -d$

From equation 2 putting the of a + dn

$a_n = 23 - d$

$a_n = 23 - 3 = 20$

Nth term is  20

Hence, the common difference and the first term of AP are 3 and 20 respectively.