The 10th term of an AP is -37 and sum of its first 6 terms is -27. Find the sum of its first eight terms.

- 63

Step-by-step explanation:

An arithmetic progression is such that the N-th element is equal to:

An = A1 + (n-1)*R

where A1 is the first term, and R is the difference between two consecutive terms.

We know that

A1 + 9*R = -37

and

A1 + (A1 + R) +  (A1 + 2R) +   (A1 + 3R) +  (A1 + 4R) +  (A1 + 5R) = -27

6*A1 + R(1 + 2 + 3 + 4 + 5) = -27

6*A1 + 15*R = -27

so we have two equations:

6*A1 + 15*R = -27

A1 + 9*R = -37

Let's find A1 and R.

First, isolate A1 in the second equation:

A1 = -37 - 9*R

now replace it in the other equation and solve it for R.

6*A1 + 15*R = -27

6*(-37 - 9*R) + 15*R = -27

-222 - 54*R + 15*R = -27

-39*R = -27 + 222 = 195

R = 195/-39 = -5

Now we can find the value of A1:

A1 = -37 - 9*R = -37 - 9*-5 = 8

So now we want to calculate the sum of the first eight terms. we already know that the sum of the first six is -27, so we need to add the seventh and the eigth.

A7 = 8 + 6*-5 = -17

A8 = 8 + 7*-5 = -22

Sum = -27 -17 - 22 = -63

7,24

step-by-step explanation: