Question

the first and the last terms of an ap are 7and49 . if the sum f all it terms is 420, find its cd.

Answer 1

Sum of all terms= n[a+l]÷2
420=n×56÷2
420=n×28
15=n

Again,
an=a+(n-1)d
49=7+14d
42=14d
3=d

Hence, the common difference of the AP is 3

Answer 2

Let the given AP contain n terms.

Here,

a = 7 , L = 49 and Sn = 420.



Therefore,

Sn = n/2 ( a + l )



=> n/2 ( 7 + 49 ) = 420



=> n/2 × 56 = 420



=> 28n = 420


=> n = 15.

Thus , the given AP contain 15 terms and T15 = 49.


Let D be the common difference of the given AP.


Then,

T15 = 49


a + 14d = 49


14d = 42

d = 3.


Hence,

The common difference of the given AP is 3.