Question

Find the sum of first 15 terms of each of the following sequences having nth term as

1. an = 3+4n
2. bn = 5+2n
3. xn = 6-n
4. yn = 9-5n.

Fast .

Answer 1

Answer:

1). Sum of first 15 term of the given sequence is 885.

2). Sum of first 15 term of the given sequence is 495.

3). Sum of first 15 term of the given sequence is -120.

4). Sum of first 15 term of the given sequence is -915.

Step-by-step explanation:

We need to find Sum of first 15 terms of the given nth term of the sequences.

1).

Given: $a_n=3+4n$

for 1st term, n = 0

we get, a = 3

for common difference,d = $a_2-a$

d = 3 + 4(2) - 3 = 8

So, Sum of first 15 term is $S_{15}$

$S_n=\frac{n}{2}(2a+(n-1)d)$

$S_{15}=\frac{15}{2}(2(3)+(15-1)8)=\frac{15}{2}(6+112)=15(59)=885$

Therefore, Sum of first 15 term of the given sequence is 885.

2).

Given: $b_n=5+2n$

for 1st term, n = 0

we get first term, a = 5

for common difference,d = $b_2-a$

d = 5 + 2(2) - 5 = 4

So, Sum of first 15 term is $S_{15}$

$S_n=\frac{n}{2}(2a+(n-1)d)$

$S_{15}=\frac{15}{2}(2(5)+(15-1)4)=\frac{15}{2}(10+56)=15(33)=495$

Therefore, Sum of first 15 term of the given sequence is 495.

3).

Given: $x_n=6-n$

for 1st term, n = 0

we get first term, a = 6

for common difference,d = $x_2-a$

d = 6 - 2 - 6 = -2

So, Sum of first 15 term is $S_{15}$

$S_n=\frac{n}{2}(2a+(n-1)d)$

$S_{15}=\frac{15}{2}(2(6)+(15-1)(-2))=\frac{15}{2}(12-28)=15(-8)=-120$

Therefore, Sum of first 15 term of the given sequence is -120.

4).

Given: $y_n=9-5n$

for 1st term, n = 0

we get first term, a = 9

for common difference,d = $y_2-a$

d = 9 - 5(2) - 9 = -10

So, Sum of first 15 term is $S_{15}$

$S_n=\frac{n}{2}(2a+(n-1)d)$

$S_{15}=\frac{15}{2}(2(9)+(15-1)(-10))=\frac{15}{2}(18-140)=15(-61)=-915$

Therefore, Sum of first 15 term of the given sequence is -915.

Answer 2

Answer:

Step-by-step explanation:

Hope you understood.