Question

The sum of 3rd and 9th of an arithmetic progression is 8.Find the sum of the first 11 terms of the progression

Answer 1

Answer:

44

Step-by-step explanation:

T3 + T9 = a+(3-1)d + [a + (9-1)d]

S = 2a + 10d = 8 ....(1)

S10 = (11/2)[2a+(11-1)d]

= (11/2)[8] (....from 1)

= 11*4 = 44