Finding the sum of the first 10 terms within an Arithmetic Progression

Question

I have been attempting to solve the following question, however, I am unable to form any sort of relationship between the two facts. I have attempted at using all three equations, however am unable to solve anything:

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

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

All the progress that I've done is to form these two equations as I am a bit stuck on how to proceed:

$T_1 + T_4 = 16$

$T_3 + T_8 = 4$

Find the sum of the first $10$ terms within an Arithmetic Progression (AP), given that; the sum of the 1st term and 4th term $= 16$, and then the sum of the 3rd term and 8th term $= 4$

I would greatly appreciate it if someone had the time to answer this question with steps to how they arrived at a conclusion.

Thanks

Answer 1

The terms $T_1$ and $T_4$ from $T_1+T_4=16$ can be rewritten using the formula $T_n=a+(n-1)d$ and substitution of $n=1$ & $n=4$:

• $T_1=a+(1-1)d=a$
• $T_4=a+(4-3)d=a+3d$

$T_3$ and $T_8$ from $T_3+T_8=4$ can be found using $S_n=\frac {n\{2a+(n-1)d\}}2$ and substituting $n=3$ & $n=8$:

• $T_3=a+(3-1)d=a+2d$
• $T_8=a+(8-1)d=a+7d$

Hence, if we rewrite and simplify the original equations $T_1+T_4=16$ and $T_3+T_8=4$ with the new values of $T_1$, $T_4$, $T_3$ and $T_8$:

- Equation 1: $a+a+3d=16 \rightarrow 2a+3d=16$

- Equation 2: $a+2d+a+7d=4 \rightarrow 2a+9d=4$

Solve simultaneously (subtract equation 1 from equation 2) to get: $$6d=-12 \rightarrow d=-2$$

Substitute $d=-2$ into either equation 1 or 2 to get $a=11$. Use this information and substitute it into $S_{10}=\frac {10\{2a+(10-1)d\}}2$

Answer 2

One might also note that the sum requested is $ \ S_{10} \ = \ 10·\frac{2·a \ + \ (10 - 1)·d}{2} \ = \ 5·(2a \ + \ 9d) \ \ . \ $ As you already noted, $ \ T_3 + T_8 \ = \ 2a + 9d \ = \ 4 \ \ , \ $ so $ \ S_{10} \ = \ 5 · 4 \ = \ 20 \ \ . \ $ (We can check from your results for $ \ a \ $ and $ \ d \ $ that $$ \ S_{10} \ = \ 11 + 9 \ + \ \underbrace{7 + 5 + 3 + 1 + (-1) + (-3) + (-5) + (-7)}_{= \ 0} \ = \ 20 \ \ . \ ) $$