Question

What is the common difference of a 43–term arithmetic sequence where the first term is –13 and the sum is 9,374?

Answer 1

S43 = 9374 = (2(13) + (42)d)(43/2)

Answer 2

Each term adds that common difference once, if there are 43 terms, then that difference has been added 42 times.
-13 + 42D = 9374
where D is the common difference

Answer 3

No Smooth. The total is 9374, not the last term

Answer 4

Ah. Indeed.

Answer 5

Then in that case,
((-13+42D) + -13)*43/2 = 9374

Answer 6

so D = 11

Answer 7

Sum 'S' of an AP of 'n' terms with the first term 'a' and common difference 'd' is:

\[\large S = \frac n 2 [2a + (n-1)d]\]


So, plug in the values of 'S', 'n' and 'a', calculate 'd' and enjoy!

Answer 8

-13.
((-13+42D) + -13)*43/2 = 9374
gives
(42D -26)*43/2 = 9374
42d-26 = 9374*(2/43)
42D -26= 436
42D = 462
D = 462/42 = 11

Answer 9

how do i plug this one in "What is the sum of a 14–term arithmetic sequence where the last term is 30 and the common difference is –5?"

Answer 10

The formula for the sum is:
[(First term) + (Last term) ]* (number of terms divided by 2)
You're given the last term. You can calculate the first term by (First term) -(D*n-1) where n is the number of terms. Plug in and solve.

Answer 11

812.5?

Answer 12

in the equation i wrote above, the part withing the brackets can be written as

...[a + a+(n-1)d]

here, if you notice, a+(n-1)d is actually the last term 'l'. so you can rewrite the whole formula as:

\[\large S= \frac n 2 (a+l)\]

Answer 13

okay when its given that the last term is given, like for example
a, b, c, d, ..... , y,z, 25

reverse the series and write

25, y, z, ...., d, c, b, a


So now the first term becomes 25. And ofcourse the common difference which was 'd' earlier, becomes '-d' (since the AP has been reversed -if it was an increasing one originally it is now decreasing, and vice-versa). And the no. of terms 'n' remains the same.

Now do the calculation for this. and then adjust for the original AP accordingly.