For an A.P . if first term a=1 and the common difference =3 then find S20
Answers
Answer:
(i) If a
n
=3−4n, show that a
1
,a
2
,a
3
,… form an A.P. Also find S
20
From the question it is given that,
n
th
term is 3+4n
So, a
n
=3−4n
Now, we start giving values, 1,2,3,… in the place of n, we get,
a
1
=3−(4×1)=3−4=−1
a
2
=3−(4×2)=3−8=−5
a
3
=3−(4×3)=3−12=−9
a
4
=3−(4×4)=3−16=−13
So, The numbers are −1,−5,−9,−13,…
Then, first term a=−1,
common difference d=−5−(−1)=−5+1=−4
We know that,
S
20
=(n/2)(2a+(n−1)d)
=(20/2)((2×(−1))+(20−1)×(−4))
=10(−2+(19×(−4)))
=10(−2−76)
=10(−78)
=−780
(ii) Find the common difference of an A.P. whose first term is 5 and the sum of the first four terms is half the sum of the next four terms.
From the question, it is given that, First term a=5
And also it is given that, the sum of first four terms is half the sum of next four terms,
a
1
+a
2
+a
3
+a
4
=1/2(a
5
+a
6
+a
7
+a
8
)
then,
a+(a+d)+(a+2d)+(a+3d)=1/2((a+4d)+(a+5d)+(a+6d)+(a+7d))
a+a+d+a+2d+a+3d=1/2(a+4d+a+5d+a+6d+a+7d)
4a+6d=1/2(4a+22d)
By cross multiplication,
2(4a+6d)=(4a+22d)
2((4×5)+6d)=((4×5)+22d)
…[ given a=5] 2(20+6d)=(20+22d)
40+12d=20+22d
40−20=22d−12d
20=10d
d=20/10
d=2
Therefore, the common difference d is 2.
★GIVEN★
- First term of AP (a) = 1
- Common Difference (d) = 3
★To Find★
The sum of 20th term.
★SOLUTION★
We know that,
where,
- n is the respective term
- a is the first term
- d is the common difference
- S is the sum of the respective term
Now,
where,
- n = 20
- a = 1
- d = 3
Putting the values,