Question

Tn=3+4n find AP and 15th term and sum of n terms​

Answer 1

Answer:

Given nth term is = 4n+1,

So 1st term is put n=1,

a1 = 4*1+1=5

And 2nd term is put n=2,

a2 = 4*2+1=9,

So difference of ap is a2-a1= 9–5 = 4,

Sum of AP is calculated by,

Sn = n/2{2a+(n-1)*d}

So sum of 15 term is,

S15 = 15/2{2*5+(15–1)*4}

S15 = 15/2{10+14*4}

S15= 15/2{10+56}

S15 = (15/2)*66

S15= 15*33

S15= 495…

Ans.. is 495

Answer 2

Answer:

ANSWER

T

n

=5n−3

T

1

=5(1)−3=2

T

2

=5(2)−3=7

d=T

2

−T

1

=7−2=5

a=T

1

=2

S

20

=

2

20

[2a+19d]

=10(4+95)

=990.