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Worked Examples · Example 3

Q.Find the number of identical terms in the two sequences: 1,5,9,13,17,…,1971, 5, 9, 13, 17, \ldots, 197 and 1,4,7,10,13,…,1961, 4, 7, 10, 13, \ldots, 196.

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The terms common to two A.P.s themselves form an A.P. whose common difference is the LCM of the two original common differences; counting its terms up to the smaller sequence's last value gives the answer.

If two A.P.s share a first term aa and have common differences d1,d2d_1,d_2, their common terms form an A.P.: a,a+D,a+2D,…a, a+D, a+2D,\ldots where D=lcm(d1,d2)D=\mathrm{lcm}(d_1,d_2). The nnth term of an A.P. is an=a+(n−1)da_n=a+(n-1)d.

  1. Sequence 1: 1,5,9,13,…,1971,5,9,13,\ldots,197; here a=1a=1, d1=4d_1=4. Number of terms: a+(n−1)d1=197⇒1+4(n−1)=197⇒n−1=49⇒n=50a+(n-1)d_1=197 \Rightarrow 1+4(n-1)=197 \Rightarrow n-1=49 \Rightarrow n=50 terms.
  2. Sequence 2: 1,4,7,10,…,1961,4,7,10,\ldots,196; here a=1a=1, d2=3d_2=3. Number of terms: 1+3(n−1)=196⇒n−1=65⇒n=661+3(n-1)=196 \Rightarrow n-1=65 \Rightarrow n=66 terms.
  3. Both sequences share the same first term 11, so their common terms form a new A.P. starting at 11 with common difference D=lcm(4,3)=12D=\mathrm{lcm}(4,3)=12.
  4. Common-term A.P.: 1,13,25,…1, 13, 25, \ldots, i.e. the kkth common term (starting k=0k=0) is 1+12k1+12k.
  5. This term must not exceed the smaller of the two last terms, min⁡(197,196)=196\min(197,196)=196: solve 1+12k≤196⇒12k≤195⇒k≤16.25⇒kmax⁡=161+12k \le 196 \Rightarrow 12k \le 195 \Rightarrow k \le 16.25 \Rightarrow k_{\max}=16.
  6. So k=0,1,2,…,16k=0,1,2,\ldots,16, giving 16−0+1=1716-0+1=17 common terms; the last one is 1+12(16)=1931+12(16)=193.
  7. Verify 193193 lies in both sequences: in sequence 1, 1+4(m−1)=193⇒m=491+4(m-1)=193\Rightarrow m=49 (valid, ≤50\le50); in sequence 2, 1+3(m−1)=193⇒m=651+3(m-1)=193\Rightarrow m=65 (valid, ≤66\le66) ✓.
✓Final answer

There are 17\boxed{17} terms common to both sequences (the common terms themselves form the A.P. 1,13,25,…,1931,13,25,\ldots,193).

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