Question

In a progression,If Tn=2n Ray's to 2+1,then s2 is,

Answer 1

Put n=1 in (1), we get,

a1=2(1)2+1

∴a1=2+1

∴a1=3

Put n=2 in (1), we get,

a2=2(2)2+1a2

=2(4)+1

∴a2=9

Thus, d=a2−a1

∴d=9−3

∴d=6

Now, sum of n terms of A.P. is,

Sn=2n

[2a+(n−1)d]

Put n=2 in above equation, we get,

S2=22

[2a+(2−1)d]

∴S2=1[(2×3)+(1×6)]

∴S2=1[6+6]

∴S2=12

Put n=1 in (1), we get,

a1=2(1)2+1

∴a1=2+1

∴a1=3

Put n=2 in (1), we get,

a2=2(2)2+1a2

=2(4)+1

∴a2=9

Thus,d=a2−a1

∴d=9−3

∴d=6

Now, sum of n terms of A.P. is,

Sn=2n

[2a+(n−1)d]

Put n=2 in above equation, we get,

S2=22[2a+(2−1)d]

∴S2=1[(2×3)+(1×6)]

∴S2=1[6+6]

∴S2=12