Question

the first and the last term of an ap at 7 and 49 respectively if the sum of all its term is 420 find its common difference

Answer 1

sum of ap is given by
s=n/2(a+l)
420=n/2(7+49)
420=n(28)
420/28=n
60/4=n
15=n

last term of a.p is given by
l = a+(n-1)d
420 = 7+(15-1)d
413/14=d

Answer 2

SOLUTION:-

Let the Given AP Contain n terms.

Here

a= 7

l= 49

Sn= 420

Sn= n/2(a+l)

= n/2(7+49)= 420

= n/2×56=420

= 28n=420

= n= 420/28

n=15

Thus,

The Given AP contain 15 terms and T15= 49

T15= 49

a+14d=49

= 7+14d=49

= 14d=42

d=3

Hence, the common difference of the Given AP is 3.