The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34.

Solution:

Let a be the first term and d be the common difference of the AP. Then,

$a_{4}=11$

$\Rightarrow a+(4-1) d=11 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+3 d=11 \quad \ldots \ldots(1)$

Now,

$a_{5}+a_{7}=34$         (Given)

$\Rightarrow(a+4 d)+(a+6 d)=34$

$\Rightarrow 2 a+10 d=34$

$\Rightarrow a+5 d=17 \quad \ldots . .(2)$

From (1) and (2), we get

$11-3 d+5 d=17$

$\Rightarrow 2 d=17-11=6$

$\Rightarrow d=3$

Hence, the common difference of the AP is 3.