Common Terms in Arithmetic Progressions

CAT 2023 Slot 3 · QA · Easy · Algebra

Let an=46+8na_n = 46 + 8n and bn=98+4nb_n = 98 + 4n be two sequences for natural numbers n100n \le 100. Then, the sum of all terms common to both the sequences is

  1. A.

    15000

  2. B.

    14900

  3. C.

    14602

  4. D.

    14798

Answer

B

Explanation

Let common terms satisfy an=bma_n = b_m: 46+8n=98+4m    8n4m=52    2nm=13    m=2n1346 + 8n = 98 + 4m \implies 8n - 4m = 52 \implies 2n - m = 13 \implies m = 2n - 13

Since 1m1001 \le m \le 100 and 1n1001 \le n \le 100: 12n13100    142n113    7n561 \le 2n - 13 \le 100 \implies 14 \le 2n \le 113 \implies 7 \le n \le 56

The common terms correspond to n{7,8,,56}n \in \{7, 8, \dots, 56\}. Number of terms N=567+1=50N = 56 - 7 + 1 = 50.

First common term = a7=46+8(7)=102a_7 = 46 + 8(7) = 102. Last common term = a56=46+8(56)=494a_{56} = 46 + 8(56) = 494.

Sum=502(102+494)=25×596=14900\text{Sum} = \frac{50}{2} (102 + 494) = 25 \times 596 = 14900

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