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Exercise 5.2 · Q1

Q.Write the first 6 terms of the sequences whose nthn^{th} terms are given below and classify them as arithmetic progression, geometric progression, arithmetico-geometric progression, harmonic progression and none of them.

(i) 12n+1\dfrac{1}{2^{n+1}}
(ii) (n+1)(n+2)n+3(n+4)\dfrac{(n+1)(n+2)}{n+3(n+4)}
(iii) 4(12)n4\left(\dfrac12\right)^n
(iv) (−1)nn\dfrac{(-1)^n}{n}
(v) 2n+33n+4\dfrac{2n+3}{3n+4}
(vi) 20182018
(vii) 3n−23n−1\dfrac{3n-2}{3^{n-1}}
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Evaluate each formula at n=1,…,6n=1,\ldots,6, then check differences (AP), ratios (GP), reciprocals (HP), or the AP×\timesGP factoring (AGP) to classify.

Step 1. (i) an=12n+1a_n=\dfrac1{2^{n+1}}. n=1,…,6n=1,\ldots,6: 14,18,116,132,164,1128\dfrac14,\dfrac18,\dfrac1{16},\dfrac1{32},\dfrac1{64},\dfrac1{128}. Ratio of consecutive terms is always 12\dfrac12 — a GP with r=12r=\dfrac12.

Step 2. (ii) an=(n+1)(n+2)n+3(n+4)a_n=\dfrac{(n+1)(n+2)}{n+3(n+4)}. The denominator, read literally as printed, is n+3(n+4)=4n+12n+3(n+4)=4n+12. Substituting: n=1:2⋅316=38n=1:\dfrac{2\cdot3}{16}=\dfrac38; n=2:3⋅420=35n=2:\dfrac{3\cdot4}{20}=\dfrac35; n=3:4⋅524=56n=3:\dfrac{4\cdot5}{24}=\dfrac56; n=4:5⋅628=1514n=4:\dfrac{5\cdot6}{28}=\dfrac{15}{14}; n=5:6⋅732=2116n=5:\dfrac{6\cdot7}{32}=\dfrac{21}{16}; n=6:7⋅836=149n=6:\dfrac{7\cdot8}{36}=\dfrac{14}9. The differences and ratios are not constant, and the reciprocals are not in AP either — classified as none of them.

Step 3. (iii) an=4(12)na_n=4\left(\dfrac12\right)^n. n=1,…,6n=1,\ldots,6: 2,1,12,14,18,1162,1,\dfrac12,\dfrac14,\dfrac18,\dfrac1{16}. Constant ratio 12\dfrac12 — a GP.

Step 4. (iv) an=(−1)nna_n=\dfrac{(-1)^n}n. n=1,…,6n=1,\ldots,6: −1,12,−13,14,−15,16-1,\dfrac12,-\dfrac13,\dfrac14,-\dfrac15,\dfrac16. Alternating in sign with shrinking magnitude — no constant difference, ratio, or reciprocal-AP pattern — none of them.

Step 5. (v) an=2n+33n+4a_n=\dfrac{2n+3}{3n+4}. n=1,…,6n=1,\ldots,6: 57,710,913,1116,1319,1522\dfrac57,\dfrac7{10},\dfrac9{13},\dfrac{11}{16},\dfrac{13}{19},\dfrac{15}{22}. Differences shrink (not constant), ratios are not constant — none of them (it does converge to 23\dfrac23 as n→∞n\to\infty, but that is a separate idea from being AP/GP/HP/AGP).

Step 6. (vi) an=2018a_n=2018. Every term is 20182018 — a constant nonzero sequence, which is simultaneously an AP (common difference d=0d=0) and a GP (common ratio r=1r=1).

Step 7. (vii) an=3n−23n−1a_n=\dfrac{3n-2}{3^{n-1}}. n=1,…,6n=1,\ldots,6: 1,43,79,1027,1381,162431,\dfrac43,\dfrac79,\dfrac{10}{27},\dfrac{13}{81},\dfrac{16}{243}. Writing an=(3n−2)(13)n−1a_n=(3n-2)\left(\dfrac13\right)^{n-1}: the numerator 3n−2=1+3(n−1)3n-2=1+3(n-1) is an AP with a=1,d=3a=1,d=3, multiplied term-by-term by the GP (13)n−1\left(\dfrac13\right)^{n-1} — this is exactly the AGP pattern (a+(n−1)d)rn−1(a+(n-1)d)r^{n-1} with a=1,d=3,r=13a=1,d=3,r=\dfrac13.

✓Final answer

(i) GP; (ii) none of them; (iii) GP; (iv) none of them; (v) none of them; (vi) both AP and GP; (vii) AGP.

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