The 7th term of an A.P. is $32$ and its ${{13}^{th}}$ term is $62.$ Find the A.P. An arithmetic progression or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solution:

Given,

${{a}_{7}}=32$ and ${{a}_{13}}=62$

From ${{a}_{n}}-{{a}_{k}}=\left( a+nd-d \right)-\left( a+kd-d \right)$

$=\left( n-k \right)d$

where n and k represents the nth and kth terms of an A.P

${{a}_{13}}-{{a}_{7}}=\left( 13-7 \right)d=62-32=30$

$6d=30$

$d=5$

Now,

${{a}_{7}}=a+\left( 7-1 \right)5=32$

$a+30=32$

$a=2$

Hence, the A.P is $2,7,12,17,$ ……