Sequences & Series · Ordinary Level

Working Backwards from Sₙ

Ordinary Level · Class 7 · Tap NEXT to begin

Section 1 of 2

From a Sum Formula to a and d

Three handy facts

$a = T_1 = S_1$ (the first sum is the first term). $T_2 = S_2 - S_1$. $d = T_2 - T_1$.

$S_n = n^2 + 3n$. Find $a$ and $d$.

$S_1 = 1^2 + 3(1) = 4$, so $a = 4$

$S_2 = 2^2 + 3(2) = 10$

$T_2 = S_2 - S_1 = 10 - 4 = 6$

$d = T_2 - T_1 = 6 - 4 = 2$

$a = 4,\ d = 2$

$S_n = 2n^2 + 3n$. Find $a$ and $d$.

$S_1 = 2 + 3 = 5$, so $a = 5$

$S_2 = 2(4) + 6 = 14$; $T_2 = 14 - 5 = 9$

$d = 9 - 5 = 4$

$a = 5,\ d = 4$

$S_n = 3n^2 + 5n - 6$. Find $a$ and $d$.

$S_1 = 3 + 5 - 6 = 2$, so $a = 2$

$S_2 = 3(4) + 10 - 6 = 16$; $T_2 = 16 - 2 = 14$

$d = 14 - 2 = 12$

$a = 2,\ d = 12$

You try

$S_n = n^2 + 7n$. Find $a$ and $d$.

$a = S_1$, $T_2 = S_2 - S_1$.

$S_1 = 8 = a$; $S_2 = 4 + 14 = 18$; $T_2 = 18 - 8 = 10$

$d = 10 - 8 = 2$

$a = 8,\ d = 2$

You try

$S_n = n^2 + 11n$. Find $a$ and $d$.

$a = S_1$.

$S_1 = 12 = a$; $S_2 = 4 + 22 = 26$; $T_2 = 26 - 12 = 14$

$d = 14 - 12 = 2$

$a = 12,\ d = 2$

You try

$S_n = 5n^2 - 3n$. Find $a$ and $d$.

$a = S_1$.

$S_1 = 5 - 3 = 2 = a$; $S_2 = 20 - 6 = 14$; $T_2 = 14 - 2 = 12$

$d = 12 - 2 = 10$

$a = 2,\ d = 10$

Working backwards: $a = S_1$, $T_2 = S_2 - S_1$, $d = T_2 - T_1$. Class 8: quadratic sequences.