Question

The sum of the first 48 terms is 4 times the sum of the first 36 terms find the sum of the first 30 terms

Answer 1

ANSWER IS ''0''

S(48) = (48/2)(2a+47d),

S(36) = (36/2)(2a+35d).

Now S(48) = 4S(36)  

, ie., 24(2a+47d) = 4*18(2a+35d) = 72(2a+35d).

Then 2a+47d = 3(2a+35d) and

6a +105d -2a -47d = 0,

i.e.,

4a +58d = 0,

i.e.,

2a+29d = 0.

Now S(30) = (30/2)(2a+29d) = (15)(2a+29d)=0

Answer 2

Answer:

1. 0

Step-by-step explanation:

sum of first n no if a= first term and d=increasing order

sum= n/2*(2*a+(n-1)d)

first we put n=48

then put n=36

then equal ,sum of( n=48)=4*sum of (n=36)

we get relation in between a and d,2a+29d=0

then sum of (n=30)=0