Sum of an AP

We need to find the sum of common terms of three arithmetic progressions:

AP1: $$3, 7, 11, 15, \ldots, 399$$ (first term $$a_1 = 3$$, common difference $$d_1 = 4$$)

AP2: $$2, 5, 8, 11, \ldots, 359$$ (first term $$a_2 = 2$$, common difference $$d_2 = 3$$)

AP3: $$2, 7, 12, 17, \ldots, 197$$ (first term $$a_3 = 2$$, common difference $$d_3 = 5$$)

Find the common difference of terms common to all three APs.

$$\text{LCM}(4, 3, 5) = 60$$

Find the first common term.

Terms in AP1 have the form $$3 + 4k$$. Terms in AP2 have the form $$2 + 3m$$. Terms in AP3 have the form $$2 + 5n$$.

We need a number $$\equiv 3 \pmod{4}$$, $$\equiv 2 \pmod{3}$$, and $$\equiv 2 \pmod{5}$$.

From $$x \equiv 2 \pmod{3}$$ and $$x \equiv 2 \pmod{5}$$: $$x \equiv 2 \pmod{15}$$, so $$x = 2 + 15j$$.

Also $$x \equiv 3 \pmod{4}$$: $$2 + 15j \equiv 3 \pmod{4} \implies 15j \equiv 1 \pmod{4} \implies 3j \equiv 1 \pmod{4} \implies j \equiv 3 \pmod{4}$$.

So $$j = 3, 7, 11, \ldots$$ giving $$x = 47, 107, 167, 227, 287, 347, \ldots$$

Find valid terms within the range of all three APs (up to $$\min(399, 359, 197) = 197$$).

Common terms $$\leq 197$$: $$47, 107, 167$$.

Compute the sum.

$$47 + 107 + 167 = 321$$

The answer is $$\boxed{321}$$.

Find the sum of terms in AP 3, 8, 13, ..., 373 that are NOT divisible by 3. We first determine the total number of terms: since $$a = 3, d = 5, a_n = 373$$ and $$373 = 3 + (n-1)(5)$$, it follows that $$n = 75$$. The total sum of all 75 terms is $$ S = \frac{75}{2}(3 + 373) = \frac{75 \times 376}{2} = 14100 $$. To identify the terms divisible by 3, observe the sequence begins 3, 8, 13, 18, 23, 28, 33, ... and a general term is $$3 + 5(k-1) = 5k - 2$$, which is divisible by 3 when $$5k - 2 \equiv 0 \pmod{3}$$, i.e., $$2k \equiv 2 \pmod{3}$$ so $$k \equiv 1 \pmod{3}$$. Thus the indices are $$k = 1, 4, 7, ..., 73$$, giving a sub-AP with first term 3, common difference 15, and last term 363. The number of terms in this sub-AP is $$\frac{363 - 3}{15} + 1 = 25$$. The sum of these 25 terms is $$ S_3 = \frac{25}{2}(3 + 363) = \frac{25 \times 366}{2} = 4575 $$. Subtracting this from the total sum gives $$14100 - 4575 = 9525$$.

The answer is 9525.

We are given an A.P. $$a_1, a_2, a_3, \ldots$$ with common difference $$d$$, such that $$\sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5}a_1$$ and $$a_1 \neq 0$$.

The terms $$a_{2k-1}$$ for $$k = 1, 2, \ldots, 12$$ are $$a_1, a_3, a_5, \ldots, a_{23}$$. This is an A.P. with first term $$a_1$$, common difference $$2d$$, and 12 terms.

$$\sum_{k=1}^{12} a_{2k-1} = 12a_1 + 2d \cdot \frac{12 \cdot 11}{2} = 12a_1 + 132d$$

Setting this equal to $$-\frac{72}{5}a_1$$:

$$12a_1 + 132d = -\frac{72}{5}a_1$$

$$132d = -\frac{72}{5}a_1 - 12a_1 = -\frac{72 + 60}{5}a_1 = -\frac{132}{5}a_1$$

$$d = -\frac{a_1}{5}$$

Now, $$\sum_{k=1}^{n} a_k = \frac{n}{2}(2a_1 + (n-1)d) = 0$$.

Since $$a_1 \neq 0$$, we need $$2a_1 + (n-1)d = 0$$ (assuming $$n \neq 0$$).

$$2a_1 + (n-1)\left(-\frac{a_1}{5}\right) = 0$$

$$2 - \frac{n-1}{5} = 0$$

$$\frac{n-1}{5} = 2$$

$$n - 1 = 10$$

$$n = 11$$

Hence, the correct answer is Option A.