Question

Eg. The sum of the common terms of the following three arithmetic progressions.

5, 11, 17, 23, ..................., 305

3, 5, 7, 9, 11, ..................., 199

6, 11, 16, 21, ..................., 301

Answer 1

707

Answer 2

To find the sum of the common terms of the three given arithmetic progressions (APs), we follow these steps:

Step 1: Identify the properties of each AP.

• AP1: 5, 11, 17, 23, ..., 305
  • First term ($a_1$) = 5
  • Common difference ($d_1$) = 11 - 5 = 6
  • Last term ($L_1$) = 305
• AP2: 3, 5, 7, 9, 11, ..., 199
  • First term ($a_2$) = 3
  • Common difference ($d_2$) = 5 - 3 = 2
  • Last term ($L_2$) = 199
• AP3: 6, 11, 16, 21, ..., 301
  • First term ($a_3$) = 6
  • Common difference ($d_3$) = 11 - 6 = 5
  • Last term ($L_3$) = 301

Step 2: Determine the common difference of the AP formed by common terms.

The common difference of the AP of common terms ($D$) is the Least Common Multiple (LCM) of the individual common differences.

$D = \text{LCM}(d_1, d_2, d_3) = \text{LCM}(6, 2, 5)$

$D = \text{LCM}(2 \times 3, 2, 5) = 2 \times 3 \times 5 = 30$

Step 3: Find the first common term.

Let the first common term be $A$. We need to find the smallest number that satisfies the conditions for all three APs.

A number $x$ is in:

• AP1 if $x \equiv 5 \pmod 6$
• AP2 if $x \equiv 3 \pmod 2$
• AP3 if $x \equiv 6 \pmod 5$

From $x \equiv 3 \pmod 2$, $x$ must be an odd number.

From $x \equiv 6 \pmod 5$, $x \equiv 1 \pmod 5$.

So, $x$ must be an odd number ending in 1 or 6. Since it's odd, it must end in 1.

Numbers ending in 1: 1, 11, 21, 31, 41, 51, 61, 71, ...

Let's check these numbers against the first two congruences:

• $x \equiv 5 \pmod 6$
• $x \equiv 1 \pmod 5$

For $x=1$: $1 \pmod 6 = 1 \ne 5$.

For $x=11$: $11 \pmod 6 = 5$ (satisfies AP1). $11 \pmod 5 = 1$ (satisfies AP3).

Also, check AP2: $11 \pmod 2 = 1$ (satisfies AP2 since $3 \pmod 2 = 1$).

Thus, the first common term ($A$) is 11.

Step 4: Determine the upper limit for the common terms.

The common terms cannot exceed the smallest of the last terms of the three APs.

Maximum common term $L = \min(L_1, L_2, L_3) = \min(305, 199, 301) = 199$.

Step 5: Find the number of common terms.

The common terms form an AP with first term $A=11$ and common difference $D=30$.

Let the last common term be $T_n = A + (n-1)D$.

We know $T_n \le L$, so $11 + (n-1)30 \le 199$.

$(n-1)30 \le 199 - 11$

$(n-1)30 \le 188$

$n-1 \le \frac{188}{30}$

$n-1 \le 6.266...$

Since $n$ must be an integer, $n-1 = 6$.

Therefore, $n = 7$.

The last common term is $T_7 = 11 + (7-1)30 = 11 + 6 \times 30 = 11 + 180 = 191$.

Step 6: Calculate the sum of the common terms.

The common terms form an AP with $n=7$ terms, first term $A=11$, and last term $L_{common}=191$.

The sum $S_n = \frac{n}{2}(A + L_{common})$

$S_7 = \frac{7}{2}(11 + 191)$

$S_7 = \frac{7}{2}(202)$

$S_7 = 7 \times 101$

$S_7 = 707$

The common terms are 11, 41, 71, 101, 131, 161, 191.

Sum = $11+41+71+101+131+161+191 = 707$.