Question

the sum of n terms of an ap is 3 n square + 5 n find the AP and find its 15 term

Answer 1

Sn = 3n² + 5n

Sn-1 = 3(n-1)² +5(n-1)
         = 3( (n)² + (1)² - 2(n)(1) ) +5n -5
         = 3(n² +1 -2n ) + 5n -5
         = 3n² + 3 - 6n +5n - 5
        =  3n² -n -2 

An = Sn - S(n-1)               
     = 3n² + 5n  - (3n² -n -2 )
    = 3n² + 5n - 3n² +n +2 
   = 6n  +2

A1 = 6(1) +2
     = 6 + 2
    = 8 

A2 = 6(2) +2
      = 12 +2 
      = 14 
d  = A2 -A1
    = 14 -8 
     = 6
AP will be
8 , 14 ,20 , 26 ...

A15 = a + (15-1)d
          = 8 +14(6)
         = 8 + 84
             = 92 
         

Answer 2

Sn = 3n² + 5n.
n = 1
Then S1 = 8 = a = 1st term.
n = 2
Then S2 = 22 = a + a2.
a2 = 14.
Therefore d = 14-8 = 6.

AP : 8,14,20,26...................

a15 = a + 14d = 8 + 14(6).
a15 = 92.

Where a = 1st term , a2 = 2nd term ,
d = common difference.


Feel free to message me if you have any doubts.

Wishing you all a HAPPY AND PROSPEROUS NEW YEAR.

Please mark mine as brainliest.