Question

For an A.P . if first term a=1 and the common difference =3 then find S20​

Answer 1

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.

Answer 2

★GIVEN★

• First term of AP (a) = 1
• Common Difference (d) = 3

★To Find★

The sum of 20th term.

★SOLUTION★

We know that,

$\large{\green{\underline{\boxed{\bf{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}}}}}$

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,

$\large\implies{\sf{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}}$

where,

• n = 20
• a = 1
• d = 3

Putting the values,

$\large\implies{\sf{S_{20}=\dfrac{20}{2}[2\times1+(20-1)3]}}$

$\large\implies{\sf{S_{20}=\dfrac{\cancel{20}}{\cancel{2}}[2\times1+(20-1)3]}}$

$\large\implies{\sf{S_{20}=10[2+19\times3]}}$

$\large\implies{\sf{S_{20}=10[2+57]}}$

$\large\implies{\sf{S_{20}=10\times59}}$

$\large\therefore\boxed{\bf{S_{20}=590}}$

Therefore,

Sum of the 20th term is 590.