Question

For an A.P if t4 =64 and t7=32,find a, d and tn​

Answer 1

Answer :-

a = 95.6

d = -10.6

an = 106.2 -10.6n

Given :-

$t_4 = 64$

$t_7 = 32$

To find:-

The value of a , d and tn.

Solution:-

Let a be the first term and d be the common - difference.

$t_4 = 64$

→$a + 3d = 64------eq.1$

$t_7 = 32$

→$a + 6d = 32------eq.2$

Subtract equation1.and equation 2,

→$a + 3d - (a + 6d) = 64 -32$

→$a -a +3d -6d = 32$

→$-3d = 32$

→$d = \dfrac{-32}{3}$

→$d = -10.6$

Now,

→a + 6d = 32

→a + 6×(-10.6)= 32

→a -63.6 = 32

→a = 32 +63.6

→a = 95.6

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

→$a_n = 95.6 + (n-1)-10.6$

→$a_n = 95.6 -10.6n +10.6$

→$a_n = 106.2 -10.6n$

Answer 2

Answer:

10.03

Step-by-step explanation:

a+3t=64

a+6t=32

_________

0-3t=32

________

t=32/3

=-10.6