Question

First term of an arithmetic sequence is 28 and common difference is -4.
a. Write the arithmetic sequence.
b .find its 8th term.
c. What is the sum of first 15 terms of the arithmetic sequence -28,-24,-29.....
please answer the question correctly....I will Mark you as brainlist​

Answer 1

Answer:

a.28,24,20,...

b.0

c.60

Step-by-step explanation:

an+b= -4n+32

c.-28,-24,-20...

s=n/2(2a+(n-1)d)

15/2(2×-28+(15-1)4)

15/2 (10)

15×5=60

hence sum= 60

a-first term

s-sum

n-number of terms

d-commom difference

Answer 2

a. A.P. series is 28, 24, 20, 16, .....

b. T8=0

C. S15=60

Step-by-step explanation:

First term of A.P is 28, and difference = -4

a. A.P series is, a, a+d, a+2d, a+3d....

28, 28+(-4), 28+2(-4), 28+3(-4)

28, 28-4, 28-8, 28-12

28, 24, 20, 16

b. a=28, d=-4

Tn= a+(n-1)d

n=8

So, T8=28+(8-1)-4

T8=28+7×-4

T8=28-28=0

T8=0

c. Given A.p series is -28, -24, -29

So, a=-28, d=4

Now we find the value of $S_{15}$.

we know that $S_n= \frac{n}{2}[2a+(n-1)d]$

$S_15= \frac{15}{2} [2(-24)+(15-1)4]\\S_15=\frac{15}{2} [-48+14(4)]\\S_15=\frac{15}{2} [-48+56]\\S_15=\frac{15}{2} [8]=60$

So, $S_15=60$

Hence, the answer is 60.

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