Question

7th term of an Ap is 40. the sum of its 13th term is ?​

Answer 1

Answer:

tn = a +(n-1)d

t7 = a + (7-1)d

40= a + 6d.......(1)

Now,

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

S13= 13/2[2a+ (13-1)d]

= 13/2[ 2a + 12d]

= 13/2×2 (a+ 6d)

= 13 (a + 6d)

Now ,

from 1 equation

40 = a +6d

S13= 40 × 13

= 520

Hope it help you.