Two APs: s1=t1≠0, s2=2t2 — Find (s2−s1)/(t2−t1)

A general problem-solving technique video — not a specific exam question.

Problem

Let $s_1,s_2,s_3,\ldots$ and $t_1,t_2,t_3,\ldots$ be two arithmetic sequences such that $s_1=t_1\neq0$ and $s_2=2t_2$. The sum of the first 10 terms of the first AP equals the sum of the first 15 terms of the second AP. Find the value of $\dfrac{s_2-s_1}{t_2-t_1}$.

Key insight. $\dfrac{s_2-s_1}{t_2-t_1}$ is just the ratio of the two common differences — since $s_2-s_1$ is the first AP’s common difference, and $t_2-t_1$ is the second’s. So the whole problem reduces to finding $d_1$ and $d_2$ (the two common differences) from the given conditions, without ever needing the actual first term.

Watch this explained step by step →  ·  More on the Shiwam’s Classes channel

Approach

Let the common first term be $A$, with common differences $d_1$ for the $s$-sequence and $d_2$ for the $t$-sequence. Translate $s_2=2t_2$ and the sum condition into two equations in $A$, $d_1$, $d_2$, eliminate $A$ between them, and solve for the ratio $d_1/d_2$ — which is exactly the quantity being asked for.

Solution

Step 1 — Set up the common first term and differences

Let $s_1=t_1=A$ (with $A\neq0$), $d_1$ = common difference of the $s$-sequence, $d_2$ = common difference of the $t$-sequence. Then $s_2=A+d_1$ and $t_2=A+d_2$.

Step 2 — Apply the condition s2 = 2t2

$$A+d_1 = 2(A+d_2) \implies A+d_1 = 2A+2d_2 \implies d_1-2d_2 = A \quad \text{…(i)}$$

Step 3 — Apply the sum condition

$$\underbrace{\frac{10}{2}\big[2A+9d_1\big]}_{\text{sum of 10 terms of }s} = \underbrace{\frac{15}{2}\big[2A+14d_2\big]}_{\text{sum of 15 terms of }t}$$

$$5(2A+9d_1) = 15(A+7d_2)$$

$$10A+45d_1 = 15A+105d_2$$

$$45d_1-105d_2 = 5A \implies 9d_1-21d_2 = A \quad \text{…(ii)}$$

Step 4 — Eliminate A between (i) and (ii)

Substituting $A=d_1-2d_2$ from (i) into (ii):

$$9d_1-21d_2 = d_1-2d_2$$

$$8d_1 = 19d_2$$

$$\frac{d_1}{d_2} = \frac{19}{8}$$

Step 5 — Relate this to the required ratio

$$\frac{s_2-s_1}{t_2-t_1} = \frac{(A+d_1)-A}{(A+d_2)-A} = \frac{d_1}{d_2} = \frac{19}{8}$$

Answer

Common mistakes

• Missing that $s_2-s_1$ and $t_2-t_1$ are simply the common differences. It’s tempting to substitute the full expressions for $s_2,s_1,t_2,t_1$ and simplify from scratch, when recognising the ratio as $d_1/d_2$ immediately connects it to the two equations already being solved.
• Losing track of which sum belongs to which sequence. The problem states the sum of 10 terms of the first (s) sequence equals the sum of 15 terms of the second (t) sequence — swapping these numbers changes every subsequent equation.

Practise next

• Two arithmetic sequences $u_1,u_2,\ldots$ and $v_1,v_2,\ldots$ satisfy $u_1=v_1\neq0$ and $u_3=3v_3$. If the sum of the first 6 terms of $u$ equals the sum of the first 9 terms of $v$, find $\dfrac{u_3-u_1}{v_3-v_1}$, using the same elimination approach.