CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If the sum of first 4 terms of an AP is 40 and that of first 14 terms is 280.  Find the sum of its first n terms.

 OR

 Find the sum of the first 30 positive integers divisible by 6.


Solution

Let a and d respectively be the first term and the common difference of the given A. P

The sum of first four terms
S4=40
42{2a+(41)d}=40
2a+3d=20(1)
The sum of first 14 terms
S14=280
142{2a+(141)d}=280
2a+13d=40  (2)
Subtracting equation (1) from equation (2)
(2a+13d)-(2a+3d) = 40 - 20
10d = 20
d = 2
Substituting d = 2 in equation (1),
Substituting d = 2 in equation (1),
2a + 3 × 2 = 20
2a = 20 - 6 = 14
a142=7
Sum of first n terms, Sn=n2{2a+(n1)d}
=n2{2×7+(n1)×2}
=n2{14+2n2}
=n2(2n+12)
=n2×(n+6)
=n(n+6)
=n2+6n

OR

The first 30 integers divisible by 6 are 6 , 12, 18 ......180

Sum of first 30 integers
= 6 + 12 + 18 +....  + 180

=302(6+180)    [Sn=n2(a+1)]
=15×186
=2790

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image