Question

The sum of the first 8 terms of an arithmetic sequence is 136 and the sum of the first 12 terms is 300. a) What is the sum of the first and the 8th terms? b) What is the sum of the first and the 12th terms ? c) What is the number got by adding three times the first term to the 19th term ?​

Answer 1

s40=

3420

Explanation:

We know that

sn=n2(2a+(n−1)d)

Thus

186=

122(2a+(11d)

186=6(2a+11d)

31=2a+11d

Now we know that tn=a+(n−1)d .

83=a+(20−1)d

83=a+19d

We now have a system of equations:

{31=2a+11d

83=a+19d

Substituting (2) into (1), we get

31=2(83−19d)+11d

31=166−38d+11d−135=−27d

d=5

Now solving for

a :83−19(5)=−12

The sum is once again given by

s40=402(2(−12)+(39)5)

s40=20(171)

s40=3420

Hopefully this helps!

Answer 2

Answer:

a)76 b)92 c)168

Step-by-step explanation:

sum of Ap = n/2(2a+(n-1)d)

putting in values

S8= 8a+28d=136....equ 1

S12= 12a+66d=300....equ 2

solving simultaneously

a=24,d=4

a)T8= a+7d=52

T8+A= 52+24=76

b)T12= a+11d=68

T12+A=92

c)3A=72

T19= A+18d=96

72+96=168