In an A. P the term is -10 and the 15th term is 11 and the last term is 41. Find the sum of all terms in this progression.​

This is equivalent to the fraction 1085/2

Explanation:

AP stands for "arithmetic progression", which is another name for "arithmetic sequence"

a1 = -10 is the first term

the 15th term happens when n = 15, so

an = a1 + d*(n-1)

a15 = -10 + d(15-1)

a15 = 14d-10

Set this equal to 11 (the stated 15th term) and solve for d

a15 = 11

14d-10 = 11

14d = 11+10

14d = 21

d = 21/14

d = 3/2

d = 1.5 is the common difference

Let's find the nth term

an = a1 + d(n-1)

an = -10 + 1.5(n-1)

an = -10 + 1.5n - 1.5

an = 1.5n - 11.5

The last term is 41, so we'll replace the 'an' with that and solve for n

an = 1.5n - 11.5

41 = 1.5n - 11.5

41+11.5 = 1.5n

52.5 = 1.5n

1.5n = 52.5

n = (52.5)/(1.5)

n = 35

So the 35th term is 41.

We're summing n = 35 terms from a1 = -10 to an = 41

S = sum of the first n terms of arithmetic progression

S = (n/2)*(a1 + an)

S = (35/2)*(-10+41)

S = 542.5

The 35 terms add up to 542.5 which is the final answer

As an improper fraction, this converts to 1085/2

step-by-step explanation: