Question

The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference and 6th term.

Answer 1

$\blue{\huge{\boxed{\star\: Solution \: \star}}}$


===> a17 =a + (17-1)d
===> a17=a +16d

===> a10= a + (10-1)d

===> a10 = a + 9d

$\green{ \boxed{according \: to \: question}}$

==> a17 = a10 + 7

===>a17 - a10 = 7

==>(a+16d) - (a+9d)=7

==> a + 16d - a -9d = 7

===> 7d=7

===>
$d = \frac{7}{7}$
===> d = 1


$\red{ \boxed{common \: difference \: is \: 1}}$

Answer 2

Solution :

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

It is given       $a_{17} - a_{10} = 7$

⇒ (a + 16d) - (a + 9d) = 7

⇒                          7d = 7  

⇒                            d = 1

Thus, the common difference is 7.