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Exercise 5.3 · Q5

Q.Find the general term and sum to nn terms of the sequence 1,43,79,1027,…1,\dfrac43,\dfrac79,\dfrac{10}{27},\ldots.

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Identify the numerator as an AP and the denominator as a GP (an AGP overall), then apply the finite-AGP-sum formula directly.

Step 1. Identify the general term. Numerators 1,4,7,10,…1,4,7,10,\ldots are an AP with a=1,d=3a=1,d=3: numerator =3n−2=3n-2. Denominators 1,3,9,27,…1,3,9,27,\ldots are a GP with ratio 33: denominator =3n−1=3^{n-1}. So

an=3n−23n−1=(1+3(n−1))(13)n−1— an AGP with a=1,d=3,r=13.a_n = \frac{3n-2}{3^{n-1}} = \big(1+3(n-1)\big)\left(\frac13\right)^{n-1} \quad\text{— an AGP with } a=1,d=3,r=\frac13.

Step 2. Apply the AGP sum formula.

Sn=a−(a+(n−1)d)rn1−r+dr(1−rn−1(1−r)2),a=1, d=3, r=13.S_n = \frac{a-(a+(n-1)d)r^n}{1-r}+dr\left(\frac{1-r^{n-1}}{(1-r)^2}\right), \qquad a=1,\ d=3,\ r=\frac13.

Step 3. Compute the first piece. 1−r=231-r=\dfrac23.

1−(3n−2)(13)n23=32[1−3n−23n].\frac{1-(3n-2)\left(\frac13\right)^n}{\frac23} = \frac32\left[1-\frac{3n-2}{3^n}\right].

Step 4. Compute the second piece. dr=3⋅13=1dr=3\cdot\dfrac13=1; (1−r)2=49(1-r)^2=\dfrac49.

1⋅1−(13)n−149=94[1−13n−1].1\cdot\frac{1-\left(\frac13\right)^{n-1}}{\frac49} = \frac94\left[1-\frac1{3^{n-1}}\right].

Step 5. Add and simplify.

Sn=32−32⋅3n−23n+94−94⋅13n−1.S_n = \frac32-\frac32\cdot\frac{3n-2}{3^n}+\frac94-\frac94\cdot\frac1{3^{n-1}}. …

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