Question

the nth term of an AP is given TN=8n+3. find the sum of first n terms of the AP.

Answer 1

Given:

Sn=8n+3.

a=S1

S2=a+a+d

=2a+d

Subtracting them we get: S2-S1=2a+d-a

=a+d =2nd term

S2-S1=a+d......................(1)

Given

Sn=8n+3

Putting n=1 and n=2 gives:-

S1=8+3

=11

a=S1=11

S2=8*2+3=16+3=19

S2-S1=19-11

=8

From(1)S2-S1=a+d

So :

a+d=8.

but a=11

so 11+d=-3

d=-3

Sn=n/2*[2a+(n-1)d]

Sn=n/2*[22-3(n-1)]

=n/2*[22-3n+3]

=n/2[25-3n]

=25n/2-3n²/2

Hope it helps you.

Answer 2

at n= 1
t1 = 8(1) +3 = 11

so, a=11

t2 = 19 = a2

so, d = 8

so, Sn. = n/2(2a+(n-1)d)

=. n/2(14+8n)
so, Sn = 7n+4n² ans.