If the last term of an AP of 30 terms is 119 and the 8th term . . .

Question : If the last term of an AP of 30 terms is 119 and the 8th term from the end (towards the first term) is 91, then find the common difference of the AP. Hence, find the sum of all the terms of the AP.

Doubt by CBSE

Solution : 

n=30

l=a₃₀=119

Now, 
8th term from the end of the AP
a₈=91
l-(8-1)d=91 [∵an=l-(n-1)d]
119-7d=91
119-91=7d
28=7d
d=28/7
d=4

Also,
a_n=a+(n-1)d
119=a+(30-1)(4)
119=a+29(4)
119=a+116
119-116=a
3=a
a=3

Now,
S_n=(n/2)[a+an]
S₃₀=(30/2)[3+119]
S₃₀=15[122]
S₃₀=1830

Alternate Method : 

n=30

a₃₀=119

a_n=a+(n-1)d
a₃₀=a+(30-1)d
119=a+29d — (1)

Now, 

8th term from the end of the AP will be (30-8+1)th i.e.23rd term from the starting so 

a₂₃=91

a_n=a+(n-1)d
a₂₃=a+(23-1)d
91=a+22d — (2)

Solving equation (1) and (2)

119=a+29d

 91=a+22d  

-     -  - 
-----------------
28=0-7d
-----------------

28=7d
28/7=d
4=d
d=4

putting in equation (2)
91=a+22(4)
91=a+88
91-88=a
3=a

a=3

Now, Sum of all the 30 terms 

S_n=(n/2)[a+an]
S₃₀=(30/2)[3+119]
S₃₀=15[122]
S₃₀=1830