Answer:
1
+t
3
+t
5
=39
t
2
+t
4
+t
6
=51
To find:
t
10
=?
Solution:
Let a be the first term and d be the common difference.
t
1
+t
3
+t
5
=39
or, a+a+2d+a+4d=39
or, 3a+6d=39
or, a+2d=13 .............(i)
t
2
+t
4
+t
6
=51
or, a+d+a+3d+a+5d=51
or, 3a+9d=51
or, a+3d=17
Subtracting eqn. from eqn. we get,
d=4
Putting value d=4 in eqn.(i) we get,
a+2×4=13
or, a=13−8
or, a=5
Now, t
10
=a+9d
t
10
=5+9×4
t
10
=5+36
t
10
=41
