Question

if sn = 3n^2+5n find Ap and its 16 term ​

Answer 1

then sum of n-1 terms= 3(n-1)2+5(n-1)

So nth term of A.P will be Tn=3n2+5n- 3(n-1)2-5(n-1)= 3n2+5n-3n2-3+6n-5n+5=6n+2

in this way by putinng values of n= 1,2,3,4,5,6,7,_______n-1, n

we will get the AP as 8,14,20,26,32,38,44,____and the 16th term will be= 6*16+2=98

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Answer 2

Sn=3n^2+5n

S1 will be equal to the first term

S1=3(1)^2+5(1)

S1=3+5

S1=8-------------- a1=8

S2=3(2)^2+5(2)

S2=3(4)+10

S2=12+10=22

When you subtract S1 from S2, you will get the second term

S2-S1=22

S2-S1=22-8

S2-S1=14-------------a2=14

AP: a, a+d, a+2d, a+3d............

We have,

a1=8

a2=14

d=a2-a1

14-8=6

d=6

The required AP is 8, 14, 20, 28, 34.........

16th term:

a16=a+(16-1)d

8+15(6)

8+90

a16 =98

The 16th term is 98.