Find the following sums :
a) 1+2+3+4+___+30
b) 8+16+24+32+__+240
c) 9+17+25+___+241
Solution and formula.
Answers
Step-by-step explanation:
Given:-
a) 1+2+3+4+___+30
b) 8+16+24+32+__+240
c) 9+17+25+___+241
To find:-
Find the following sums ?
Solution:-
a) Given that
1+2+3+4+----+30
First term = 1
Common difference
= 2-1 = 1
= 3-2 = 1
Since the common difference is same throughout the series.
They are in the Arithmetic Progression.
Now we have
First term = a = 1
Common difference =d = 1
last term = an = 30
We know that
an = a+(n-1) d
=> 30 = 1+(n-1)1
=> 30= 1+n-1
=> n = 30
Number of terms in the given AP = 30
We know that
Sum of n terms in an AP
=> Sn = (n/2)[a+an)
=> S30= (30/2)(1+30)
=> S30 = 15(31)
=> S30= 465
b) a) Given that
8+16+24+32+---+240
First term = 8
Common difference
= 16-8 = 8
= 24-16 = 8
Since the common difference is same throughout the series.
They are in the Arithmetic Progression.
Now we have
First term = a = 8
Common difference =d = 8
last term = an = 240
We know that
an = a+(n-1) d
=> 240 = 8+(n-1)8
=> 240 = 8+8n-8
=>240 = 8n
=> 8n = 240
=> n = 240/8
=> n = 30
Number of terms in the given AP = 30
We know that
Sum of n terms in an AP
=> Sn = (n/2)[a+an)
=> S30= (30/2)(8+240)
=> S30 = 15(248)
=> S30= 3720
c) Given that
9+17+25+---+241
First term = 9
Common difference
= 17-9 = 8
= 25-17 = 8
Since the common difference is same throughout the series.
They are in the Arithmetic Progression.
Now we have
First term = a = 9
Common difference =d = 8
last term = an = 241
We know that
an = a+(n-1) d
=> 241 = 9+(n-1)8
=> 9+8n-8=241
=>8n+1 = 241
=> 8n = 241-1
=> 8n = 240
=>n = 240/8
=> n = 30
Number of terms in the given AP = 30
We know that
Sum of n terms in an AP
=> Sn = (n/2)[a+an)
=> S30= (30/2)(9+241)
=> S30 = 15(250)
=> S30= 3750
Answer:-
a) 1+2+3+4+----+30 = 465
b)8+16+24+32+---+240=3720
c)9+17+25+---+241 = 3750
Used formulae:-
- an = a+(n-1)d
- Sn = (n/2)[a+an]
- an = nth term or last term or General term
- n = Number of terms
- d = Common difference
- a=First term