seventeenFOREVER.: Qs of Progression!

I Hate AMath. [4.5/365] by vulnerablechaos.

$T_3=3, S_8=0$

Question: find the sum of first 50th term and find the sum of the terms from the 11st term to 50th term

Remember the key formula?

$T_n = a + (n-1)d$

$S_n = \frac{n}{2}(2a + (n-1)d)$

First, use the information given

$T_3 = 3, a + (3-1)d = 3$

$S_8 = 0, \frac{8}{2}(2a+7d) = 0$

Solve it one by one

$\frac{8}{2}(2a+7d) = 0$

$4(2a+7d) = 0$

$8a+28d = 0$

$a = -28d/8, a = \frac{-7d}{2}$

now we got a here $a = \frac{-7d}{2}$

substitute a into the term equation $a+2d = 3$

$\frac{-7d}{2}+2d = 3$

$-7d+4d = 6$

$-3d = 6$

$d = -2$

now we got $d = -2$ , to find a just substitute d back into the term equation $a+2d = 3$ , and voila! we got a!

$a = 7$

here your way to solve the first question become easy! use the information that you get.

$a = 7,d=-2$

then substitute into $S_{50}$

$S_{50} = \frac{50}{2}(2(7)+49(-2))$

$S_{50} = -2100$

now we move to the second question. This question is quite tricky to solve. To find the eleventh to fiftieth term. Wow! 11 to 50? er.. Nevermind just find it!

It’s quite tricky here on the eleventh term. We will use the 10th term instead of eleventh term. Why? Because we need the eleventh term to include in the sum that we wanted to find. If you use the eleventh term to find the sum of 11th to 50th term, the sum that you find will exclude the eleventh term.

This is the concept.

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