Question

the nth term of an AP is given by Tn=4n+3. find the sum of first 60 terms of the AP.

Answer 1

T1=4(1)+3=7

T60=4(60)+3=243

Sn=n/2(a+Tn)
S60 = 30(7+243)
S60= 7500

HOPE IT HELPS

Answer 2

Given Tn=4n+3

Putting n=1 and 2 in this equation we get:

T1=4*1+3

=4+3=7

T1=7..........(1)

T2=4*2+3

=8+3=11...............(2)

Let the first term be a

Let common difference be d.

a is equal to T1.

a=7.[from(1)]

a+d=T2

T2=11[See(1)

So a+d=11

7+d=11

d=11-7=4

S60=60/2*[2a+(n-1)d]

=30*[14+59*4]

=30[14+236]

=30*250

=7500.

The answer is 7500

Hope it helps you.

Have a nice day ahead :-)